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TC-Practice-Record

rikisand — Sat, 26 Dec 2009 09:28:00 GMT

tchs-1 none 1000pt DFS 利用�q�入的方向划分四个边

tchs-2 250pt 直接��就�?我写�?�?500pt 暴力可以�q�，但是判断时候不能用stringstream 用算术判�?也可以用构造法 1000pt 每一位有三种可能�?/p>

不用�Q�保持不动，变化�Q�分别递归计算value�q�更新结果即可，�׃��递归深度最多只�?3层所以不会tle

另外也可以写出基��Cؓ3的��@环来遍历每一�U�情况具体看代码

for(i=0,A[0]++;A[i]>2;i++){
A[i]=0;A[i+1]++;
}

tchs-3 1000pt 要想使乘�U�最大，需要更多的3卛_�� 500pt 又看错题�?~~~ft 要注意题目一定要看清�?/p>

tchs-4 500pt 模拟题，好难�?音乐的~ 可以都乘�?6 用整数来计算 ��点会很烦~ �q�种题思�\要清�?一步一步来

rikisand 2009-12-26 17:28 发表评论

TOPCODER-SRM455

rikisand — Sat, 19 Dec 2009 06:07:00 GMT

摘要: 杯具的比赛~�W�一题竟然把南和北写反了- -�Q�第二题没判断好复杂度，实际上暴力方法可以的�Q�第三题dp 必然没写出来 so----跌成灰色了~~ 250pt Problem Statement Petya likes spiders. He put a spider in each cell of a rectangular grid. He has studied spiders for many ... 阅读全文

rikisand 2009-12-19 14:07 发表评论

TC-SRM-454_div2

rikisand — Sun, 13 Dec 2009 02:18:00 GMT

摘要: 200pt( �? 500pt Problem Statement You are playing a game with a NxN grid of letters. The goal of the game is to spell out a N-letter word somewhere on the grid either horizontally from left to right o... 阅读全文

rikisand 2009-12-13 10:18 发表评论

Topcoer SRM 453.5

rikisand — Fri, 11 Dec 2009 11:59:00 GMT

摘要: 一直没写，补上上次的srm~ 250pt 500pt 队列BFS Code Snippet template void getMin(T& a,T b){if(b void getMax(T& a,T b){if(b>a)a=b;} #define REP(i, n)... 阅读全文

rikisand 2009-12-11 19:59 发表评论

TC-SRM-233-1000pt BFS

rikisand — Wed, 25 Nov 2009 03:39:00 GMT

�l�箋是misof 数字教学里面的习题~ �W�一��的最后一道题了~

Problem Statement

You are in a maze containing revolving doors. The doors can be turned 90 degrees by pushing against them in either direction. You are to find a route from the start square to the end square that involves revolving as few doors as possible. Given a map of the maze, determine the fewest number of door revolutions necessary to get from the start to the end.

In the map:

   ' ': empty space
   '#': wall
   'O': center of a revolving door (letter "oh", not zero)
   '-': horizontal door (always adjacent to a 'O')
   '|': vertical door (always adjacent to a 'O')
   'S': start square
   'E': end square

Each revolving door will always be oriented horizontally (with two horizontal segments) or vertically (with two vertical segments):

    |
    O  or  -O-
    |

Doors can be revolved 90 degrees by moving onto a door segment from any of the 4 squares diagonally adjacent to the door center, i.e., the 'X' characters below:

   X|X     X X
    O  or  -O-
   X|X     X X

Here is an example map:

        ###
        #E#
       ## #
    ####  ##
    # S -O-#
    # ###  #
    #      #
    ########

In this example, 2 door revolutions are necessary to get from 'S' to 'E'. The first turn is shown here:

        ###         ###
        #E#         #E#
       ## #        ## #
    ####  ##    #### |##
    # S -O-#    # S  OX#
    # ### X#    # ###| #
    #      #    #      #
    ########    ########

And the second turn is shown here:

        ###         ###
        #E#         #E#
       ## #        ## #
    ####X|##    #### X##
    # S  O #    # S -O-#
    # ###| #    # ###  #
    #      #    #      #
    ########    ########

Your method should return an int, the minimum number of door revolutions necessary to get from the start square to the end square. If there is no way to reach the end square, return -1.

Definition

Class:
RevolvingDoors

Method:
turns

Parameters:
vector

Returns:
int

Method signature:
int turns(vector map)

(be sure your method is public)

Notes

-
Assume that all squares off the edge of the map are walls.

Constraints

-
map will contain between 3 and 50 elements, inclusive.

-
Each element of map will contain between 3 and 50 characters, inclusive.

-
Each element of map will contain the same number of characters.

-
Each character in map will be one of 'S', 'E', 'O', '-', '|', '#', and ' ' (space).

-
There will be exactly one 'S' and one 'E' in map.

-
There will be between 1 and 10 doors, inclusive, and they will be formatted in map as described in the problem statement.

-
No two doors will be close enough for any part of them to occupy the same square.

-
It is not allowed for a door to be blocked and unable to turn. There will not be any walls in any of the 4 squares immediately adjacent to the center of a door, nor will a door be on the edge of the map.

关键是门的状态表�C�，参考了�|�站上的代码�Q�挑了一个比较简�l�的�Q�用的优先��队列。写完调好发现TLE 囧~ copy出网站上的再run依然TLE``` ``` 看了下发现现在的system testing 比原来增加了几个��试用例~ 于是扑և�misof大牛的解法，发现对状态处理一��L(f��ng)��~只不�q�用了memo和deque�Q�省��M��优先�U�队列调整的旉��开销�Q�改好了��pass了~ 上代码~~�Q?/strong>

Code Snippet

using namespace std;
typedef long long int64;
typedef vector<int> VI;
typedef vector VS;
#define inf 1000000
#define REP(i,n) for(int (i)=(0);((i)<(n));++(i))
template<class T> inline void checkmin(T &a,const T &b) { if (b
template<class T> inline void checkmax(T &a,const T &b) { if (b>a) a=b; }
int dr[]={-1,0,1,0};
int dc[]={0,1,0,-1};
struct state{state(int x,int y,int z,int s):r(x),c(y),doorstate(z),best(s){}int r;int c;int doorstate;int best;};
int memo[56][56][1<<11];
class RevolvingDoors
{
        public:
        int turns(vector mp)
        {
             int x=mp.size()+2;int y=mp[0].size()+2;
             int sr,sc,er,ec,cnt=0,doorInit=0;
             REP(i,x-2)mp[i]='#'+mp[i]+'#';                //trick:modify the pattern to make it easy to loop
             mp.insert(mp.begin(),string(58,'#'));
             mp.push_back(string(58,'#'));
             REP(i,x)REP(j,y)if(mp[i][j]=='S'){mp[i][j]=' ';sr=i;sc=j;}else if(mp[i][j]=='E'){mp[i][j]=' ';er=i;ec=j;}
             REP(i,x)REP(j,y)if(mp[i][j]=='O'){if(mp[i-1][j]=='|')doorInit|=(1<
                mp[i-1][j]=mp[i+1][j] = 100 + cnt*2 +1;    //use the content in the box to identify the door number,and the door pos
                mp[i][j-1]=mp[i][j+1] = 100 + cnt*2 ;    //if pos==0 it means this box is on the left or right of the door
                cnt++; mp[i][j]='#';
             }
             REP(i,x)REP(j,y)REP(t,1<//init the memo
             deque Q; Q.push_back(state(sr,sc,doorInit,0));
             while(!Q.empty()){
                state now=Q.front();Q.pop_front();
                int r=now.r  , c=now.c  , door=now.doorstate , b=now.best;
                if( memo[r][c][door] < b)continue;    //no better ,no vist
                REP(dir,4){                            //try four direction
                    int nr=r+dr[dir],nc=c+dc[dir];
                    if(mp[nr][nc]==' '){
                        if(memo[nr][nc][door] > b){ memo[nr][nc][door]=b;Q.push_back(state(nr,nc,door,b));}
                    }
                    else if(mp[nr][nc]=='#')continue;
                    else{                            //if we come to a box near to the door-mid
                        int doornum=(mp[nr][nc]-100)/2;int open=(mp[nr][nc]-100)%2;
                        if( ((door>>doornum)&1) != open){    //lucily,the box is empty
                            if(memo[nr][nc][door] > b){memo[nr][nc][door] = b;Q.push_back(state(nr,nc,door,b));}
                        }
                        else {                                // the box has a door
                            if(open==0 && dr[dir]==0) continue;    //try to define the relative pos between the direction and the box
                            if(open==1 && dc[dir]==0) continue;    //also ~ if we cannot push the door we give up
                            int ndoor=door^(1<//we can go into the box if we push the door ~
                            if(memo[nr][nc][ndoor] > b+1 ){memo[nr][nc][ndoor] = b+1 ;Q.push_back(state(nr,nc,ndoor,b+1));}
                        }
                    }
                }
             }
             int ans=inf;
             REP(i,1<//loop to check the best ans~
                 if(memo[er][ec][i]" "<" "<
             }
             if(ans == inf) return -1;
             else return ans;
        }

中文copy是�ؕ�?#183;··囧啊~~ 俺的破烂英文注释啊~~~

rikisand 2009-11-25 11:39 发表评论

TCCC-05-1000pt-

rikisand — Mon, 23 Nov 2009 07:29:00 GMT

Problem Statement

Just before a chess match between two teams, each team's coach secretly determines an ordering of his team's players. The first players in each team then get to play against each other, the second players in each team play against each other, etc. The team with the most wins will win the match.
You are the coach for one of the teams, and you have somehow managed to find out the order of the players in the other team. Based on that, you want to order your players so that your team's expected score is maximized to your advantage. The expected score of a single game can be calculated by the following formula (which is directly derived from how the international chess rating system works):
E_A = 1 / (1 + 10 ^{(R_B - R_A)/400}) E_B = 1 / (1 + 10 ^{(R_A - R_B)/400})

where R_A and R_B are the ratings of player A and B, and E_A and E_B are the expected scores for each player. For instance, if R_A is 2432 and R_B is 2611, the expected score for player A is 1 / (1 + 10^179/400) = 0.263005239459. The expected score for a team is the sum of the expected scores for the players in that team.
To make things a bit more complicated, the players in a team must be ordered such that a player with rating x plays before all players with rating strictly less than x - lim, where lim is a given non-negative integer. For example, if lim is 100, a player with rating 2437 must play before a player with rating 2336 but not necessarily before a player with rating 2337.
Create a class ChessMatch containing the method bestExpectedScore which takes a vector teamA, the ratings of your players (in no particular order); a vector teamB, the ratings of your opponent's players in the order they will play; and an int lim, the limit determining how freely you can order your players. You can assume that your opponent's players will be ordered in accordance with lim. The method should return a double, your team's expected score if you order your players optimally.
Definition

Class:
ChessMatch
Method:
bestExpectedScore
Parameters:
vector , vector , int
Returns:
double
Method signature:
double bestExpectedScore(vector teamA, vector teamB, int lim)
(be sure your method is public)

要求最好的得分�Q�要遍历所有的排列方式昄��不可能。如果�{化问题�ؓ依次攄��每个选手到每一个位�|�，那么只要求出剩下选手排列中��得得分最大的��可以了�Q�可以想到dp�Ҏ(gu��)��的最有子问题定义�?/font>
用每一位记录当前的选手分配情况�Q�如果�ؓ0�Q�则已经分配�Q�如果�ؓ1�Q�则�q�未分配。pos来记录当前要分配的位�|�。对于每一个子集，考虑每一个�ؓ1的也��是未分配的选手使位于pos位置�Q�然后check 是否满��在他之后的球员（也就是剩下的�?的位�Q�是否满��x��件的不等式，满��则更新当前分配的最优成�l�。最�l?2^N-1 的成�l�即为所求�?/font>

Code Snippet

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define two(X) (1<<(X))
#define contain(S,X) ((S&two(X))>0)
#define eps 1e-9
double p[20][20];
int N;
double memo[1<<20];
int lim;VI A;
bool check(int now,int pos){
    REP(i,N)if(now&two(i)){
        if(A[pos]+limreturn false;
    }
    return true;
}
double solve(int now,int pos){
    if(now==0) return 0;
    if(memo[now]>-eps)return memo[now];
    REP(i,N) if(contain(now,i)&&check(now^two(i),i)){
         double tmp=p[i][pos]+solve(now^two(i),pos+1);
         if(tmp>memo[now]+eps) memo[now]=tmp;
    }
    return memo[now];
}

class ChessMatch
{
        public:
        double bestExpectedScore(vector <int> tA, vector <int> tB, int _lim)
        {
            N=tA.size(); lim=_lim;A=tA;
            REP(i,N)REP(j,N)  p[i][j]=1.0/(1.0+pow(10.0,double(tB[j]-tA[i])/400.0));
            REP(i,1<
            return solve(two(N)-1,0);
        }

上面的方法��用递归�Q�下面是使用递推的程序。即从子集出发最�l�到�?^n-1的思�\。在�q�里把A队伍先从大到��排序。在每一个子集记录每一个未分配选手的得分和要分配的位置�?/font>

�E�序中思�\为：依次从子集中选取一个选手作�ؓ�W�bits-1个选手�Q�也��是说子集中剩下的元素要攑֜��q�个选手前面�Q�由于选手从小到大遍历�Q�只需要遍历到s[j]<=s[bits-1]+lim 为止�Q�也��p��满��q�个子集中最��的选手+lim >=选中�?卛_��Q�然后更新当前子集的得分

注释中思�\�? 依次从子集选取一个选手作�ؓ当前子集中的�W�一个元素，也就是整体的�W�n-bits个元素，�q�次从大到小遍历�Q�这样只需要保证，位于子集�W�一个的元素+lim>=子集中最大的元素(s[0])�Q�即可，然后更新当前子集得分

最�l�全�?^n-1的即为所�?/font>

Code Snippet

class ChessMatch
{
        public:
        double bestExpectedScore(vector <int> ta, vector <int> tb, int _lim)
        {
                int n=ta.size();a=ta;lim=_lim;N=n;
                sort(ta.begin(),ta.end(),greater<int>());int s[32],pos[32];
                REP(i,n)REP(j,n)p[i][j]=1./(1.+pow(10.,double(tb[j]-ta[i])/400.));
                REP(mask,1<
                    int bits=0;
                    dp[mask]=0.;
                    if(!mask)continue;
                    REP(i,n) if(contain(mask,i)){
                        s[bits]=ta[i];pos[bits++]=i;
                    }
                    //for (int j = 0 ; j < bits  && s[j] + lim >= s[0]; j++){
                    //    if(dp[mask] < p[pos[j]][n-bits]+dp[mask^two(pos[j])])
                    //        dp[mask]= p[pos[j]][n-bits]+dp[mask^two(pos[j])] ;

                    for (int j = bits-1; j >= 0 && s[j] <= s[bits-1] + lim; j--){
                    if(    dp[mask] <  p[pos[j]][bits-1] + dp[mask & ~(1 << pos[j])] )
                      dp[mask] = p[pos[j]][bits-1] + dp[mask & ~(1 << pos[j])];
                    }
                }
                return dp[two(n)-1];
        }

rikisand 2009-11-23 15:29 发表评论

TC-05TCOR4-250 DP

rikisand — Sun, 22 Nov 2009 07:58:00 GMT

roblem Statement

Character j in element i (both 0-based) of messages denotes how many messages employee i sent to employee j. You will return a vector containing the hierarchy of the employees within the company. Element 0 is the highest ranked employee, and so forth. The returned ranking should minimize the number of messages sent from lower ranked employees to higher ranked employees. If multiple solutions are possible, return the one with element 0 minimal. If there are still ties, minimize element 1, etc.
Definition

Class:
CompanyMessages
Method:
getRank
Parameters:
vector
Returns:
vector
Method signature:
vector getRank(vector messages)
(be sure your method is public)
Constraints

-
messages will contain between 2 and 15 elements inclusive.
-
Each element of messages will contain exactly N characters, where N is the number of elements in messages.
-
Each character in messages will be a digit ('0'-'9').
-
Character i in element i of messages will be '0'.
按照题目说明可知�Q�按照找到的��序�Q�所有从低��向高�U�发的信得数目和应该是最��的。又观察刎ͼ�只要��定了第一个，也就是��别最高的boss�Q�那么剩下的员工怎么排列�Q�他们向最高boss 发信数是不变的，从而子问题��是在剩下员工中扑ֈ� 低��向高�U�发信数最��的排列 �Q�显然这�W�合DP问题最优子�l�构的问题�?/font>

T[1···N]=Min{sum(N <->{1-N-1},T[1-N-1] } 得到状态方�E�后很容易写出程�?�Q�可以��用备忘录dp

Code Snippet

#define REP(i, n) for (int i = 0; i < (n); ++i)

#define two(X) (1<<(X))
#define contain(S,X) ((S&two(X))>0)

int A[1<<15];
vectorint> > vec(1<<15);
class CompanyMessages
{
        public:
        vector <int> getRank(vector mes )
        {
            int n=mes.size();
            REP(mask,two(n)){
                if(!mask)continue;
                A[mask]=100000;
                REP(i,n)if(contain(mask,i)){
                    int now=mask^two(i); int tmp=A[now];
                    REP(j,n)if(contain(now,j))tmp+=(mes[j][i]-'0');
                    if(tmp
                        vec[mask].clear();vec[mask].push_back(i);
                        for(int s=0;s
                        A[mask]=tmp;
                    }
                }
            }
            return vec[two(n)-1];
        }

Code Snippet

int memo[two(15)],record[two(15)];
VS M;int N;VI ans;
int solve(int now){
    if( now == 0 ) return 0;
    if( memo[now]>=0 )return memo[now];
    memo[now] = 100000;
    REP(i,N)if(contain(now,i)){
        int mask = now ^ two(i);int tmp = solve(mask);
        REP(j,N)if(contain(mask,j))tmp += (M[j][i] - '0');
        if(tmp < memo[now]) {
            memo[now]=tmp ;
            record[now] = mask;
        }
    }
    return memo[now];
}
void cacl(){
    int mask = two(N)-1;
    ans.clear();
    REP(i,N){
        int k = mask ^ record[mask];
        int cnt = -1; while(k>0){k>>=1;cnt++;}
        ans.push_back(cnt);
        mask=record[mask];
    }
}
class CompanyMessages
{
        public:
        vector <int> getRank(vector mes)
        {
               M=mes;N=mes.size();
               memset(memo,-1,sizeof(memo));
               solve(two(N)-1);
               cacl();
               return ans;
        }

rikisand 2009-11-22 15:58 发表评论

SRM 453 div2

rikisand — Wed, 18 Nov 2009 07:24:00 GMT

半夜12�Q?0的比赛，开始就发现�pȝ��很慢�Q�结果第一题没法提交，然后退出重�q�退出重�q?#183;····server down·····

惌��v�z�的��服�q�没拿，于是跑下��d��服�Q�看了会dota vod �Q�重新登入，比赛竟然�l�箋·····交了500 开始看1000 没思�\�Q�冻��M��?/font>

challenge没啥意思，都没人在�Q?50的太弱智都对�?#183;···�׃��中间出错�Q�这�ơsrm not rated~~

250 pt 略过

500 pt

John and Brus have an interest in team sports tournaments. They are currently investigating a basketball tournament. Basketball is a team sport in which two teams of five players try to score points against one another by placing a ball through a ten foot high hoop. Basketball is one of the most popular and widely viewed sports in the world.
There are n teams in the tournament. Each pair of teams plays exactly two games against each other. In the first of these games, one of the teams is the host, and in the second, the other team is the host. Each game results in one team winning. There are no draws. After the tournament is over, the team with the highest total number of wins is crowned the winner.
The tournament is currently in progress and the current results are described in the vector table. For each pair of distinct indices, i and j, the j-th character of the i-th element of tableis the result of the game where team i hosted team j. The result is 'W' if team i won, 'L' if team i lost, and '?' if the game hasn't been played yet. Assuming that every possible outcome is possible for the games that haven't been played yet, return the minimal number of total wins the tournament winner can have at the end of the tournament.
Definition

Class:
TheBasketballDivTwo
Method:
find
Parameters:
vector
Returns:
int
Method signature:
int find(vector table)
(be sure your method is public)
Constraints

-
table will contain between 2 and 5 elements, inclusive.
-
Each element of table will contain exactly n characters, where n is the number of elements in table.
-
The j-th character of the i-th element of table, where i and j are different, will be 'W', 'L', or '?'.
-
The i-th character of the i-th element of table will be 'X'.

数据量很��，扑ֈ�未比的比赛场��?�Q�然后枚丑֐��U�输赢情况，更新解就可以了。或者��@�?^n�ơ，或者递归调用

1000 pt

n this tournament, each game results in either a victory for one team or a draw. If a team wins a game, it gains three points and its opponent gains no points. In case of a draw, each team gains one point. The score of a team is the sum of all the points it has gained from all its games. Each pair of teams can play against each other any number of times.
You are given a vector points representing the current standings in the tournament. The i-th element of points is the score of the i-th team. You can assume that the points represent a valid state, i.e., intermediate standings that can be achieved in a tournament according to the rules described above.
Each team will play exactly one more game in the tournament, but it is not known what the matchups will be. After the tournament is over, the teams will be ranked by score. 1st place will go to the team with the highest score, 2nd place will go to the team with the second highest score, and so on. If two teams have the same score, the team with the lower number will place higher. For example, if team 0 and team 3 each have the highest score of 100 points, then team 0 will place 1st and team 3 will place 2nd.
John's favorite team is team 0, and he wants it to place as high as possible. Assuming that the remaining games can be scheduled arbitrarily and can end with any possible outcome, return the highest possible place for team 0 at the end of the tournament.
Definition

Class:
TheSoccerDivTwo
Method:
find
Parameters:
vector
Returns:
int
Method signature:
int find(vector points)
(be sure your method is public)
Constraints

-
points will contain between 2 and 50 elements, inclusive.
-
points will contain an even number of elements.
-
Each element of points will be between 0 and 1,000,000, inclusive.
-
points will represent a valid state.
Examples

Code Snippet

int find(vector <int> p )
{
     int E3=0,A3=0,L3=0,L=0;
     int d=p[0];
     for(int i=1;i
        if(p[i]>d+3)A3++;
        else if(p[i]==d+3)E3++;
        else if (p[i]>d)L3++;
        else L++;
     }
     if(A3+L+1>=E3)
         return A3+1;
     return A3+1+(E3-A3-L)/2;
}

因�ؓ每队只剩下一场比赛，所以题目变得很��单�?队最后一轮肯定是取到3分，�?队多3��Z��上比赛的肯定�?队靠前，�?队分��或者相�{�的一定在0队之后，剩下的就是我们要考虑的了。如果A3+L+1>=E3 也就是说�?队多胜一场的队伍�Q�如果让他们在最后一轮都输，那么0队可以获得最好成�l?A3+1;

然而如果不行剩下的�q�些E3要有一�?E3+1)/2个要得到3分从而比0队高�Q�over~

rikisand 2009-11-18 15:24 发表评论

rikisand — Thu, 12 Nov 2009 14:03:00 GMT

昨天看了sedgewick的alg in c 的第一章，主要介绍了个find-union 判断联通的��法�Q�其实用��C��q�查集的知识

刚才偶然看见poj�q�查集的题集 ��copy�q�来自己也写写看~~·

慢慢来~~ 写完一道上一道~~

       POJ 1182 食物�?/strong>
http://acm.pku.edu.cn/JudgeOnline/problem?id=1182

�q�以为是最��单的一道呢�Q�被列在最前面�Q�原来是扩展的应用，不是那么直接啊~

�U�结了一会儿�Q�看了解题报告才有思�\。关键是在�ƈ查集的基��上增加一个数�l�，�l�护其与根节点的关系�Q?/font>

详情写到注释里面吧：
int A[100003][3]; int root[50005],k[50005];//root记录节点所属于的树木id k记录节点与根的关�p? 同类1子是食物2子是天敌 int sz[50005]; int find(int x){ if(root[x]==x)return x;//如果是根则返�? else if(root[root[x]]==root[x])return root[x];//如果�W�一层，则返�? int father=root[x]; root[x]=find(father);//攑ֈ��W�一层，减少下次find旉��Q�减��树的高�? k[x]=(k[x]+k[father])%3;//��p��׃��根的关系以及父节点与根的关系推断出新的自�׃��根的关系�Q�实际上是�ؓ了解决union后的节点与根的关�p?/span>
//因�ؓunion后，非根节点�q�没有更新k�Q�因此需要在此处处理更新 return root[x]; } inline void Union(int x,int y,int kind){ int a=find(x);int b=find(y); if(sz[a]>sz[b]){//判断后决定，要把size��的攑ֈ�size大的上去�Q�优化作用不�? root[b]=a ; //b成�ؓa的子�? k[b]=(k[x]-k[y]+kind+3)%3;//要把b变成a的子树，从a出发�Q�到x�Q�经�q�kind�Q�到y�Q�再到b �Q�可以得到kb的状态改变公�? sz[a]+=sz[b];} else{ if(kind==1)kind=2; root[a]=b; k[a]=(k[y]-k[x]+kind+3)%3; sz[b]+=sz[a]; } } int main() { int N,d; freopen("d:\\input.txt","r",stdin); scanf("%d %d",&N,&d); for(int i=1;i<=N;i++)sz[i]=1; int cnt=0; for(int s=1;s<=N;s++)root[s]=s; int LIE=0; int i=0; while(i !=d) { scanf("%d %d %d",&A[i][0],&A[i][1],&A[i][2]); if(A[i][1]>N || A[i][2]>N ) {LIE++;i++;continue;} if(A[i][0]==1){ if(find(A[i][1]) == find(A[i][2])){ if(k[A[i][1]]!=k[A[i][2]]) LIE++; } else Union(A[i][1],A[i][2],0); } else{ if( find(A[i][1])==find(A[i][2]) ){if(k[A[i][2]] != ( k[A[i][1]]+1)%3)LIE++;} else Union(A[i][1],A[i][2],1); } i++; } cout<return 0; }
POJ 1611 The Suspects

http://acm.pku.edu.cn/JudgeOnline/problem?id=1611

int G[30005]; int sz[30005]; int Find(int x){ if(G[x]==x)return x; else { while(G[x]!=x){//half路径压羃~�{�下试试看全压羃的效率~那样��是递归调用�? int t=G[x]; G[x]=G[t]; x=t; } return x; } } void Union(int x,int y){ if(sz[x]int main() { freopen("d:\\input.txt","r",stdin); int N,M,num; while(true){ scanf("%d %d",&N,&M); if(N==0&&M==0)return 0; for(int i=0;i1; for(int i=0;i"%d",&num); int root,stu,x,y; for(int j=0;j"%d",&stu); if(j==0)root=Find(stu);//��单的都和�W�一个合�q? else{ root=Find(root); x=Find(stu);if(root!=x)Union(root,x); } } } printf("%d\n",sz[Find(0)]); } }

POJ 2524 Ubiquitous Religions

又一道水题~~ 呵呵

不脓(chu��ng)代码�?和上面一道基本一样~
http://acm.pku.edu.cn/JudgeOnline/problem?id=2524

POJ 1861

http://acm.pku.edu.cn/JudgeOnline/problem?id=1861

kruskal+�q�查�?half路径压羃

比较基础的题
struct Edge{ int from; int to; int value; }E[15000]; int A[1005]; int sz[2009]; bool comp(const Edge& a,const Edge& b){ return a.valueusing namespace std; int Find(int x){ if(A[x]==x)return x; else if(A[A[x]]==A[x]) return A[x]; int father; while(A[x]!=x){ father=A[x]; A[x]=A[father];//把每一个�\�q�的节点攑ֈ��父下面�? x=father; } return x; } void Union(int x,int y){ if(sz[y]>sz[x])swap(x,y);//��的攑ֈ�大的下面 sz[x]+=sz[y]; A[y]=x; } int main() { freopen("d:\\input.txt","r",stdin); int N,M,num,x,y; scanf("%d %d",&N,&M); int cnt=0; while(cnt"%d %d %d",&E[cnt].from,&E[cnt].to,&E[cnt].value),cnt++; sort(E,E+M,comp);//从小到大选n-1条边�Q�如果�v点和�l�点在一个集合则continue�Q�else Union for(int i=0;i<=N;i++)sz[i]=1,A[i]=i; vector<int> ans(N-1); int mst=0,MX=0; for(int i=0;mst!=N-1;i++){ int x=Find(E[i].from);int y=Find(E[i].to); if(x==y)continue; Union(x,y); ans[mst]=i; mst++; if(E[i].value>MX)MX=E[i].value; } printf("%d\n%d\n",MX,N-1); for(int i=0;i1;i++) printf("%d %d\n",E[ans[i]].from,E[ans[i]].to); }

        POJ 1456 Supermarket
http://acm.pku.edu.cn/JudgeOnline/problem?id=1456

       POJ 1733 Parity game
http://acm.pku.edu.cn/JudgeOnline/problem?id=1733

       hdu 3038 How Many Answers Are Wrong
http://acm.hdu.edu.cn/showproblem.php?pid=3038

     POJ 1417 True Liars(�?
http://acm.pku.edu.cn/JudgeOnline/problem?id=1417

   POJ 2912 Rochambeau(�?
http://acm.pku.edu.cn/JudgeOnline/problem?id=2912

rikisand 2009-11-12 22:03 发表评论

rikisand — Thu, 12 Nov 2009 13:45:00 GMT

Your restaurant has numTables tables to seat customers. The tables are all arranged in a line. If a large party of customers comes in, a group of adjacent tables will be used. Which group of tables is entirely up to the customer. Since you cannot predict this, assume all possible choices occur with equal probability. What you can predict is the size of each group of customers that arrives. Element i of probs gives the probability, in percent, that an entering party will need i+1 tables. Assuming nobody leaves, return the expected number of tables you will use before a party must be turned away. This only occurs if there is no place to seat them.
Method signature:
double getExpected(int numTables, vector probs)

numTables will be between 1 and 12 inclusive.
probs will contain between 1 and 12 elements inclusive.
Each element of probs will be between 0 and 100 inclusive.
The elements of probs will sum to 100.

misof 数字表达教程里的习题~ 题目大意求��用桌子的期望。由于到来group的个��C��定，每个group需要的桌子不定�Q��ɼ��定期望变得困难。但考虑对于numTables来说,使用桌子的状态仅仅有 2^numTables�U�，因此考虑在这些状态改变的�q�程中来计算期望�Q�也��是计算在每个状态下面的期望桌子数目。在每个状态到达时�Q�依�ơ考虑来了一个group需要k个位子，如果r�U�安排可以满��k个位子，那么当前状态的期望��D��加上来k个位子的概率 X �Q�r�U�安排分别的期望�?/ r�Q?其中求r中安排期望和则需�?递归调用函数。显然利用memo可以减少重复计算于是有下面的解法�Q?/font>
vector p;
double dp[1<<13];
int tb;
double solve(int cur){
    if(dp[cur]>-1.0)return dp[cur];    //memo available
    double ret=0;double sum;int kind;
    for(int i=0;i        sum=0,kind=0;
        int mask=(1<<(i+1))-1;    //new group need i+1 adjacent tables
        for(int j=0;j+i+1<=tb;j++){
            if((cur&(mask<                sum+=solve(cur+(mask<                kind++;
            }
        }
        if(kind!=0)sum/=kind; //caculate the average need
        ret+=sum*p[i];
    }
    dp[cur]=ret;
    return ret;
}
        double getExpected(int numTables, vector probs)
        {
                tb=numTables;
                REP(i,1<<13)dp[i]=-1.0;
                p.resize(probs.size());
                for(int i=0;i                return solve(0);//the beginning pattern
        }
看比赛中有另一�U�解法，��x��据题目，在到达每�ơfail to serve a group 的时�?�Ҏ(gu��)��此时的桌子数量，和到达这�U�状态的概率来计��：

dp[1<<13][15];memset(dp,0,sizeof(dp));// :D lucily I can do this for 0

double fails=0.0;bool flag ;

for(int i=1;i<=numTables+1;i++) //循环最多numTables+1 ��?/font>

{flag=true;

for(int j=0;j

     int mask=(1<<(j+1))-1;//注意�U�M��q�算�W�的优先�U�低,注意加括�?/font>

     for(int k=0;k<=(1<

          if(dp[k][i-1]<=0.0)continue;

          flag=false;

          int cnt=0;

          for(int m=0;m+j+1<=numTables;m++) if((mask<

          if(cnt)for(int m=0;m+j+1<=numTables;m++)if((mask<

          if(!cnt){

                 int b=k,bn=0;while(b){if(b&1)bn++;b>>=1;}

                 fail+=dp[k][i-1]*bn;

         }

    }

}

if(flag)return fail;//all dp[][k]==0.0

}

return fail;

优先�U�很�Ҏ(gu��)��错：

http://www.cppreference.com/wiki/operator_precedence~。~

典型的几�?/font>

++ --

~ !& *&

* / %

+ -

>> <<

< <= > >=

== !=

&

^ xor

|

&&

||

?=

= += –= <<= >>=

,

从上��C��依次降低~~~~~~~~~~~~~~~~~~··

rikisand 2009-11-12 21:45 发表评论

久久国产色AV免费看,无码人妻久久久一区二区三区,91精品国产色综久久

TC-Practice-Record

TOPCODER-SRM455

TC-SRM-454_div2

Topcoer SRM 453.5

TC-SRM-233-1000pt BFS

�l�箋是misof 数字教学里面的习题~ �W�一���的最后一道题了~

Problem Statement

Definition

Notes

Constraints

TCCC-05-1000pt-

Problem Statement

Definition

TC-05TCOR4-250 DP

roblem Statement

Definition

Constraints

SRM 453 div2

Definition

Constraints

Definition

Constraints

Examples

Method signature:double getExpected(int numTables, vector probs)

�l�箋是misof 数字教学里面的习题~ �W�一��的最后一道题了~

Method signature:
double getExpected(int numTables, vector probs)