欧美jjzz,国产精品久久久久一区二区三区 ,欧美精品aa

HDOJ 1253 HDU 1253 胜利大逃亡 ACM 1253 IN HDU

MiYu — Wed, 25 Aug 2010 03:30:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋

题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=1253

题目描述:

胜利大逃亡Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4785    Accepted Submission(s): 1454


Problem Description
Ignatius被魔王抓��C��,有一天魔王出差去�?�q�可是Ignatius逃亡的好��Z��.

��王住在一个城堡里,城堡是一个A*B*C的立方体,可以被表�C�成A个B*C的矩�?刚开始Ignatius被关�?0,0,0)的位�|?���d��城堡的门�?A-1,B-1,C-1)的位�|?现在知道��王���在T分钟后回到城�?Ignatius每分钟能从一个坐标走到相�ȝ��六个坐标中的其中一�?现在�l�你城堡的地�?请你计算出Ignatius能否在魔王回来前���d��城堡(只要走到出口���q�����d��城堡,如果走到出口的时候魔王刚好回来也���逃亡成功),如果可以误���出需要多���分钟才能离开,如果不能则输�?1.


 

Input
输入数据的第一行是一个正整数K,表明���试数据的数�?每组���试数据的第一行是四个正整数A,B,C和T(1<=A,B,C<=50,1<=T<=1000),它们分别代表城堡的大���和��王回来的时�?然后是A块输入数�?先是�W?�?然后是第1�?�W?�?.....),每块输入数据有B�?每行有C个正整数,代表�q�宫的布局,其中0代表�?1代表�?(如果对输入描�q�C��清楚,可以参考Sample Input中的�q�宫描述,它表�C�的���是上图中的�q�宫)

特别注意:本题的测试数据非常大,请��用scanf输入,我不能保证��用cin能不���时.在本OJ上请使用Visual C++提交.
 

Output
对于每组���试数据,如果Ignatius能够在魔王回来前���d��城堡,那么误�����Z��最���需要多���分�?否则输出-1.
 

Sample Input
1
3 3 4 20
0 1 1 1
0 0 1 1
0 1 1 1
1 1 1 1
1 0 0 1
0 1 1 1
0 0 0 0
0 1 1 0
0 1 1 0
 

Sample Output
11
 

题目分析:

    题目是很���单的一道BFS, 但是�q�破题竟然纠�l�了我一整天, 晚上11点的时候才A�?YM.  
�q�题�?二维矩阵的不同点���是�?个方�? 前后左右上下.  然后一直BFS���可以了,不需要剪枝都能过...........

代码如下:
 /*
MiYu原创, 转帖��h���?: 转蝲�?______________白白の屋
          http://www.cnblog.com/MiYu
Author By : MiYu
Test      : 1
Program   : HDU1253
*/

#include 
#include 
using namespace std;
int TLE[56][56][56];
const int d[6][3] = { { 0,0,1 },{ 0,0,-1 },{ 0,-1,0 },{ 0,1,0 },{ 1,0,0 },{ -1,0,0 } };
int A, B, C, T, m;
typedef struct pos{
       pos(){ x=y=z=n=0; } 
       void setPos ( int a,int b,int c, int count ){ x=a;y=b;z=c;n=count; }
       bool isEnd () { if ( x==1&&y==1&&z==1 )return true;return false; }
       int x,y,z;  
       int n;
}pos;
pos t,p;
#define CMP(A,B) (A.n < B.n)
typedef class Heap {
public:
        pos h[70000 * 2];
        int n, p, c;
        Heap() {
                n = 0;
        }
        void inline push(pos e) {
                for (p = ++n; p > 1 && CMP(e,h[p>>1]); h[p] = h[p>>1], p >>= 1)
                        ;
                h[p] = e;
        }
        int inline pop(pos &e) {
                if (!n)
                        return 0;
                for (e = h[p = 1], c = 2; c < n
                                && CMP(h[c += (CMP(h[c + 1],h[c]) && c < n - 1)], h[n]);
                                h[p] = h[c], p = c, c <<= 1)
                        ;
                h[p] = h[n--];
                return 1;
        }
}Heap;
Heap WA;
int RE ()
{
    if ( A+B+C-2 > T || TLE[A][B][C] == 0 )
         return -1;
    t.setPos ( A,B,C,0 );
    TLE[A][B][C] = 0;
    WA.push ( t );
    while ( WA.pop(t) ){ 
           if ( t.x+t.y+t.z-2 > T-t.n )
                continue;            
           if ( t.isEnd() )
                return t.n;  
           for ( int i = 0; i != 6; ++ i ){
                 int x = t.x+d[i][0], y=t.y+d[i][1], z=t.z+d[i][2];
                 if ( TLE[ x ][ y ][ z ] != 0 ){
                      TLE[ x ][ y ][ z ] = 0;
                      p.setPos ( x, y, z, t.n+1 ); 
                      if ( p.isEnd() && p.n <= T )
                           return p.n;  
                      if ( p.n <= T )    
                           WA.push ( p );
                 }
           }  
    }
    return -1;
}
inline bool scan_d(int &num) 
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-'){ IsN=true;num=0;}
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
int main ()
{
    int K;
    scan_d(K);
    while ( K -- ){
           while ( WA.pop(p) ) ;
           memset ( TLE, 0 , sizeof ( TLE ) );
           scan_d(A); scan_d(B); scan_d(C); scan_d(T);
           for ( int i = 1; i <= A; ++ i ){
                 for ( int j = 1; j <= B; ++ j ){
                       for ( int k = 1; k <= C; ++ k ){
                             scan_d(m);
                             TLE[i][j][k] = m == 1 ? 0 : 1; 
                       } 
                 } 
           } 
           TLE[1][1][1] = 1;
           cout << RE () << endl;
    } 
    //system( "pause" );
    return 0;
}

MiYu 2010-08-25 11:30 发表评论

HDOJ 1181 HDU 1181 变�Ş�?ACM 1181 IN HDU

MiYu — Wed, 25 Aug 2010 03:27:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋

题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=1181

题目描述:

代码变�Ş�?br>
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2655    Accepted Submission(s): 863


Problem Description
�?.....变�Ş课上Harry���到了一点小�ȝ��,因�ؓ他�ƈ不像Hermione那样能够��C��所有的咒语而随意的���一个棒球变成刺猬什么的,但是他发��C��变�Ş咒语的一个统一规律:如果咒语是以a开头b�l�尾的一个单�?那么它的作用���恰好是使A物体变成B物体. 
Harry已经���他所会的所有咒语都列成了一个表,他想让你帮忙计算一下他是否能完成老师的作�?���一个B(ball)变成一个M(Mouse),你知�?如果他自�׃��能完成的�?他就只好向Hermione��h��,�q�且被迫听一大堆好好学习的道�?
 

Input
���试数据有多�l�。每�l�有多行�Q�每行一个单�?仅包括小写字�?是Harry所会的所有咒�?数字0表示一�l�输入结�?
 

Output
如果Harry可以完成他的作业,���p���?/span>"Yes.",否则���p���?/span>"No."(不要忽略了句�?
 

Sample Input
so
soon
river
goes
them
got
moon
begin
big
0
 

Sample Output
Yes.

题目分析:

此题是一个很标准�?nbsp;搜烦�? 直接枚�D + 回溯 ��?nbsp;OK�?nbsp;.

代码/*
MiYu原创, 转帖��h���?nbsp;: 转蝲�?nbsp;______________白白の屋
          http://m.shnenglu.com/MiYu
Author By : MiYu
Test      :
Program   :
*/

#include<iostream>
#include<string>
using namespace std;
struct{
        char beg;
        char end;
}M[101];
bool hash[101],f;
int N;
bool DFS ( char ch )
{
     if ( f )
        return true;
     if( ch == 'm' )
     {
         f = true;
         return true;
     }
     for ( int i = 0; i < N; ++ i )
         if ( M[i].beg == ch && !hash[i] )
         {
              hash[i] = true;
              DFS ( M[i].end );
              hash[i] = false;
         }
     return false;
}
int main ()
{
    string str;
    while ( cin >> str )
    {
            N = 0;
            f = false;
            memset ( hash, 0 , sizeof ( hash ) );
            while ( str != "0" )
            {
                    M[N].beg = str[0];
                    M[N].end = str[ str.size() - 1 ];
                    N++;
                    cin >> str;
            }
            DFS ( 'b' );
            puts ( f ? "Yes." : "No." );
    }
    return 0;
}

代码其实�q�题�q�有一�U�很 YD 的解�?/span>!!!  嘿嘿 ................ 
具体情况看代�?

#include<iostream>

using namespace std;

char ss[10];

int main(){

    int flag=1;

    while(gets(ss)){

        if (strcmp(ss,"0")==0){

            if (flag){

                printf("Yes.\n");

                flag=0;

            }

            else

                printf("No.\n");

        }

    }

    return 0;

}

MiYu原创, 转帖��h��?: 转蝲�?a > ______________白白の屋

MiYu 2010-08-25 11:27 发表评论

HDOJ 1016 HDU 1016 Prime Ring Problem ACM 1016 IN HDU

MiYu — Wed, 18 Aug 2010 04:32:00 GMT

题目地址:

http://acm.hdu.edu.cn/showproblem.php?pid=1016

题目描述:

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6150 Accepted Submission(s): 2745

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

Input

n (0 < n < 20).

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

Sample Input

6
8

Sample Output

Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2
题目分析:
	典型�?DFS 题目, 不需�?什么剪�? 直接 �I��D + 回溯 ���OK�? 不过值得一提的�?�q�题输出很BT,  一般的 前后 输出 回�R , �W�一个回车用
if( n == 1 )  回�R;  来做PE了好几次,  最后直接在�E�序最�?/span> 输出2个回车符竟然���A�? YM�?..........	
代码如下:
/*
MiYu原创, 转帖��h���?: 转蝲�?______________白白の屋
          http://m.shnenglu.com/MiYu
Author By : MiYu
Test      :
Program   :
*/

#include 
using namespace std;
bool prim[25];
int res[25];
bool hash[25];
int N;
void setPrim ()
{
     memset ( prim, 0, sizeof ( prim ) );
     prim[2] = prim[3] = prim[5] = prim[7] = prim[11] = prim[13] = prim[17] = prim[19] = prim[23] = true;    
}
bool DFS ( int num , int n )
{
     res[n] = num;
     if ( n > N )
     {
           return false;     
     }
     if ( n == N - 1 )
     {    
          for ( int i = 2; i <= N; ++ i )
          {
                if ( prim[num + i] && prim[ i + 1 ] && !hash[i] ) 
                {
                     res[n+1] =  i;
                     for ( int i = 1; i <= N; ++ i )
                           printf ( i == 1 ? "%d" : " %d",res[i] );
                     putchar ( '\n' );
                }
          }
     }
     for ( int i = 2; i <= N; ++ i )
     {
           if ( prim[ num + i ] && !hash[i] )
           {    
                hash[i] = true;
                DFS ( i, n + 1 );
                hash[i] = false;   
           }
     } 
     return false;     
}
int main ()
{
    setPrim ();
    int ca = 1;
    while ( cin >> N )
    {
            sizeof ( hash, 0 , sizeof ( hash ) );
            printf ( "Case %d:\n",ca++ );
            hash[1] = true;
            DFS ( 1, 1 );
            putchar ( '\n' );
    }
    return 0;
}

MiYu 2010-08-18 12:32 发表评论

HDOJ 1010 HDU 1010 Tempter of the Bone ACM 1010 IN HDU

MiYu — Wed, 18 Aug 2010 02:35:00 GMT

摘要: MiYu原创, 转帖��h��?: 转蝲�?nbsp;______________白白の屋题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=1010 题目描述: 代码Tempter of the BoneTime Limit: 2... 阅读全文

MiYu 2010-08-18 10:35 发表评论

HDOJ 1969 HDU 1969 Pie ACM 1969 IN HDU

MiYu — Wed, 11 Aug 2010 06:37:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1969
题目描述:

Pie

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 65

Problem Description
My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

Input
One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.

Output
For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).

Sample Input
3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

Sample Output
25.1327
3.1416
50.2655

题目分析:

2分求�?

题目大意是要办生日Party�Q�有n个馅��|��有f个朋友，接下来是n个馅饼的半径。然后是分馅��g��Q?br>注意咯自�׃��要，大家都要一样大�Q��Ş状没什么要求，但都要是一整块的那�U�，也就是说不能从两个饼�?br>各割一��块来凑一块，像面�U��ؓ10的和6的两块饼�Q�饼的厚度是1�Q�所以面�U�和体积相等�Q�，
如果每�h分到面积�?�Q�则10分两块，6切成5�Q�够�?个�h�Q�如果每�?�Q�则只能分两个了�Q?br>题目要求我们分到的饼��可能的大！

只要注意�_�ֺ�问题��可以了,一般WA 都是�_�ֺ�问题.

代码如下:

#include <iostream>
#include <cmath>
using namespace std;
double a[10005];
int N,f;
double pi = acos ( -1.0 );
bool test ( double x )
{
    int count = 0;
    for ( int i = 1; i <= N; ++ i )
    {
        count += int ( a[i] / x );
    }
    if ( count >= f + 1 )
    {
         return true;
    }
    else
    {
         return false;
    }
}
int main()
{
    int T;
    scanf ( "%d", &T );
    while ( T -- )
    {
           double sum = 0;
           double rad;
           scanf("%d%d",&N,&f);
           for ( int i = 1; i <= N; ++ i )
           {
               scanf ( "%lf", &rad );
               a[i] = rad * rad * pi;
               sum += a[i];
           }
           double max = sum / ( f + 1 );
           double l = 0.0;
           double r = max;
           double mid = 0.0;
           while ( r - l > 1e-6 )
           {
                  mid = ( l + r ) / 2;
                  if ( test ( mid ) )
                  {
                       l = mid;
                  }
                  else
                  {
                       r = mid;
                  }
           }
           printf("%.4lf\n",mid);
    }
    return 0;
}

MiYu 2010-08-11 14:37 发表评论

HDOJ 2899 HDU 2899 Strange fuction ACM 2899 IN HDU

MiYu — Wed, 11 Aug 2010 05:51:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2899
题目描述:

Strange fuction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 95    Accepted Submission(s): 75

Problem Description
Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)

Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

Sample Input
2
100
200

Sample Output
-74.4291
-178.8534

题目分析:

�U�数学题, 需要微�U�分的知识分析题�?..........我承�?...我数学没怎么学好..........
分析出来使用2分搜索就可以�?.........

代码如下:

#include <iostream>
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
double cal ( double n )
{
       return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
int main ()
{
    int T;
    scanf ( "%d",&T );
    while ( T -- )
    {
          scanf ( "%lf",&y );
          if ( cal(0) > y || cal(100) < y )
          {
               printf ( "No solution!\n" );
               continue;
          }
          double l = 0.0, r = 100.0,res = 0.0;
          while ( r - l > 1e-6 )
          {
                double mid = ( l + r ) / 2.0;
                res = cal ( mid );
                if ( res > y )
                     r = mid - 1e-6;
                else
                     l = mid + 1e-6;
          }
          printf ( "%.4lf\n",( l + r ) / 2.0 );
    }
    return 0;
}

MiYu 2010-08-11 13:51 发表评论

HDOJ 2199 HDU 2199 Can you solve this equation? ACM 2199 IN HDU

MiYu — Wed, 11 Aug 2010 05:29:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2199
题目描述:

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 322 Accepted Submission(s): 148

Problem Description
Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.

Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10);

Output
For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.

Sample Input
2
100
-4

Sample Output
1.6152
No solution!

题目分析:

很明�?�q�是一�?分搜索的题目, 但是注意下题目的数据!! 1e10 的实�?! 而且�_�ֺ�是要求在 0.0001 . 所以就��是2分数据量依旧比较�?如果�?br>通常的递归�Ҏ��吗很遗憾 , RE�?............ 没办�? 只能循环�?
下面的是递归 RE 的代�?:

#include 
#include 
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
bool douEql ( double a,double b )
{
if ( fabs( a - b ) <= 1e-6  )
return  true;
return false;
}
double cal ( double n )
{
return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
double biSearch ( double l, double r )
{
if ( douEql ( l,r ) )
{
if ( douEql ( y, cal ( l ) ) )
return l;
return -1;
}
double mid = ( l + r ) / 2.0;
if ( douEql ( y, cal ( mid ) ) )
return mid;
else if ( cal ( mid ) > y )
return biSearch ( l,mid - 0.0001 );
else
return biSearch ( mid + 0.0001, r );
}
int main ()
{
int T;
scanf ( "%d",&T );
while ( T -- )
{
scanf ( "%lf",&y );
if ( cal(0) >= y && cal(100) <= y )
{
printf ( "No solution!\n" );
continue;
}
double res = biSearch ( 0.0, 100.0 );
if ( res == -1 )
printf ( "No solution!\n" );
else
printf ( "%.4lf\n",res );
}
return 0;
}

AC代码如下:

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

#include <iostream>
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
double cal ( double n )
{
       return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
int main ()
{
    int T;
    scanf ( "%d",&T );
    while ( T -- )
    {
          scanf ( "%lf",&y );
          if ( cal(0) > y || cal(100) < y )
          {
               printf ( "No solution!\n" );
               continue;
          }
          double l = 0.0, r = 100.0,res = 0.0;
          while ( r - l > 1e-6 )
          {
                double mid = ( l + r ) / 2.0;
                res = cal ( mid );
                if ( res > y )
                     r = mid - 1e-6;
                else
                     l = mid + 1e-6;
          }
          printf ( "%.4lf\n",( l + r ) / 2.0 );
    }
    return 0;
}

MiYu 2010-08-11 13:29 发表评论