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chinaeli — Tue, 27 May 2008 17:36:00 GMT

#include
#include
#include
using namespace std;
struct NODE
{
int p_x,p_y,step,c_x,c_y,d;
};
queue que;
const int MAXN = 100;
bool p[MAXN][MAXN][4],c[MAXN][MAXN];
int map[MAXN][MAXN];
int mv[4][2]={-1,0,0,1,1,0,0,-1},p_sm,p_sn,c_sm,c_sn,m,n;
void init ( )
{
memset(p,false,sizeof(p));
for ( int i=0 ; i  for ( int j=0 ; j  {
   scanf("%d",&map[i][j]);
   if ( map[i][j]==4 )
    p_sm=i,p_sn=j;
   else
    if ( map[i][j]==2 )
     c_sm=i,c_sn=j;
  }
}
int bfs ( )
{
NODE temp,go;
int i;
temp.p_x = p_sm;
temp.p_y = p_sn;
temp.c_x = c_sm;
temp.c_y = c_sn;
temp.step = 0;
temp.d=0;
for ( i=0 ; i<4 ; i++ )
{
  p[p_sm][p_sn][i]=true;
  temp.d=i;
  que.push(temp);
}
c[c_sm][c_sn]=true;
while ( !que.empty () )
{
  NODE head=que.front();
  que.pop() ;
  for ( i=0 ; i<4 ; i++ )
  {
   int tm=head.p_x+mv[i][0] , tn=head.p_y+mv[i][1] ;
   if ( tm>=0 && tm=0 && tn   {
    double t=sqrt( (double)( (tm-head.c_x)*(tm-head.c_x) ) +(double)( (tn-head.c_y)*(tn-head.c_y) ));
    if ( t <= sqrt(2.0) && t>0 )
    {
     go=head;
     go.p_x = tm;
     go.p_y = tn;
     go.step ++ ;
     go.d=i;
     p[tm][tn][i]=true;
     que.push(go);
    }
    else
    {
     int tmm=tm+mv[i][0] , tnn=tn+mv[i][1];
     if ( t==0 && tmm>=0 && tmm=0 && tnn     {
      go.c_x = tmm ;
      go.c_y = tnn ;
      go.p_x = tm;
      go.d=i;
      go.p_y = tn;
      go.step = head.step+1;
      if ( map[tmm][tnn]==3 )
       return go.step ;
      c[tmm][tnn]=true;
      p[tm][tn][i]=true;
      que.push(go);
     }
    }
   }
  }
}
return -1;
}
int main ( )
{
while ( scanf("%d%d",&n,&m)!=EOF )
{
  init();
  printf("%d\n",bfs());
// que.

}
}

chinaeli 2008-05-28 01:36 发表评论

11

chinaeli — Tue, 06 May 2008 16:22:00 GMT

#include
#include
using namespace std;
class FIELDDiagrams
{
public:
  void dfs ( int k, int max, int n , long long& sum , int a )
  {
   int i;
   if ( max==n )
   {
    sum++;
    return;
   }
   if (( k-1)*a+1    return ;
   for ( i=a ; i>=1   ; i-- )
    dfs(k-1,max,n+i ,sum , i );
  }
  long long countDiagrams ( int t )
  {
   long long map[40][40],sum=0,i;
   memset(map,0,sizeof(map));
   for ( i=1 ; i<=t*(t-1)/2 ; i++ )
    dfs( t, i,0,sum, i );
   return sum;
  }
};
int main ( )
{
FIELDDiagrams a;
int n;
while (cin>>n)
  cout<}

chinaeli 2008-05-07 00:22 发表评论

chinaeli — Wed, 16 Apr 2008 15:27:00 GMT

#include"cqueue.h"
#include<iostream>
void char_queue::enqueue(char d)
{
    if ( tail>=max )
        throw ( bad_op( ) );
    else
        c[tail++]=d;
}
char char_queue::dequeue()
{
    return c[head++];
}
int char_queue::isEmpty()
{
    return head==tail?1:0;
}

#include<cstdio>
class char_queue
{
    int head,tail,max;
    char c[10000];
public:
    char_queue() {head=0; tail=0; max=10000; }
    void enqueue( char );
    char dequeue( );
    int isEmpty();
    class bad_op
    {
    public :
        int type ;
        bad_op(  ) { type=1 ; }
    };

};

chinaeli 2008-04-16 23:27 发表评论

chinaeli — Thu, 03 Apr 2008 15:10:00 GMT

#include<iostream>
#include<sstream>
#include<map>
#include<string>
#include<vector>
#include<iomanip>
#include<fstream>
using namespace std;
typedef map< string , vector<int> > WORD;
void init ( WORD & m )
{
ifstream fin;
fin.open("keywords.txt");
    string keyword;
    while ( fin>>keyword )
        m[keyword];
fin.close();
}
void count ( WORD & m )
{
    int line=0;
ifstream fin;
fin.open("text.txt");
    string str;
    while ( getline(fin,str) )
    {
        stringstream SS(str);
        line++;
        while(SS>>str)
        {
            if ( m.find(str)!=m.end() )
                m[str].push_back (line);
        }
    }
fin.close();
}
void output ( WORD m )
{
    string keyword;
    for ( WORD::iterator iter_map=m.begin() ; iter_map!=m.end() ; iter_map++ )
    {
        cout<<setw(10)<<iter_map->first<<":"
            <<"(";
        for ( vector<int>::iterator iter_vector=iter_map->second.begin() ; iter_vector!=iter_map->second.end(); iter_vector++ )
        {
            if ( iter_vector!=iter_map->second.begin() )
                cout<<",";
            cout<<*iter_vector;
        }
        cout<<")"<<endl;
    }
}
int main  ()
{
    WORD m;
    init(m);
    count(m);
    output(m);
}

chinaeli 2008-04-03 23:10 发表评论

chinaeli — Wed, 02 Apr 2008 15:21:00 GMT

#include<iostream>
#include<sstream>
#include<map>
#include<string>
#include<vector>
#include<iomanip>
#include<fstream>
#include<algorithm>
using namespace std;
typedef map< string , vector<int> > WORD;
void init ( WORD & m )
{
ifstream fin;
fin.open("keywords.txt");
    string keyword;
    while ( fin>>keyword )
        m[keyword];
fin.close();
}
void count ( WORD & m )
{
    int line=0;
ifstream fin;
fin.open("text.txt");
    string str;
    while ( getline(fin,str) )
    {
  line++;
  for ( WORD::iterator iter = m.begin () ; iter!=m.end( ); iter++ )
   if ( str.find ( iter->first )!=string::npos )
    m[iter->first].push_back(line);
    }
fin.close();
}
void only( WORD &m )
{
for ( WORD::iterator i=m.begin() ; i!=m.end(); i++ )
{
  vector<int>::iterator  new_end=unique(i->second.begin(),i->second.end());
  i->second.erase(new_end,i->second.end());
}
}
void sort_list ( WORD m , map<int,string> &mm )
{
for ( WORD::iterator iter_m = m.begin() ; iter_m != m.end() ; iter_m++ )
  mm[iter_m->second .size ()]=iter_m->first;
}
void output ( WORD m , map<int,string> mm )
{
    string keyword;
for ( map<int,string>::reverse_iterator iter=mm.rbegin() ; iter!=mm.rend() ; iter++ )
    {
        cout<<setw(10)<<iter->second<<":"
            <<"(";
  for ( vector<int>::iterator iter_vector=m[iter->second].begin() ; iter_vector!=m[iter->second].end(); iter_vector++ )
        {
            if ( iter_vector!=m[iter->second].begin() )
                cout<<",";
            cout<<*iter_vector;
        }
        cout<<")"<<endl;
    }
}
int main  ()
{
    WORD m;
map<int , string> mm;
    init(m);
    count(m);
sort_list(m,mm);
only(m);
    output(m,mm);
}

chinaeli 2008-04-02 23:21 发表评论

chinaeli — Mon, 31 Mar 2008 12:57:00 GMT

我就是不理解�q�个投票规则�Q�对�q�题整整困惑�?2个小�Ӟ��从下�?点到凌晨5炏V��。。。。。�?br>Long March Voting

Description

Instant run-off voting is a system for selecting the most preferred candidate in an election. At the beginning of the process, each voter ranks the candidates from most preferred to least preferred. A series of automated voting rounds are then held to determine the overall winner.

In each round, each voter casts a single vote for his most preferred remaining candidate. If a candidate receives strictly more than 50% of the votes cast in that round, that candidate is declared the winner of the election. Otherwise, the candidate with the fewest votes in that round is eliminated, and another round is held. If multiple candidates are tied for the least number of votes, they are all eliminated. If all the candidates are eliminated, the election ends without a winner.

You are given the preferences of the voters in an election, and you must determine the outcome. There are M candidates numbered 0 to M-1, inclusive. The preferences are given in N lines, where each element describes the preferences of a single voter. This is a permutation of the digits 0 to M-1 in decreasing order of preference. In other words, the first digit is the voter's most preferred candidate, the second digit is his second most preferred candidate, and so on.

Input

There are several test cases,each test case begins with a integer N(1<=N<=50),means there are N voters.The next N lines,each contains a string with the same lenth M(1<=M<=10).Each element of a voter will be a permutation of the digits between 0 and M-1. There is a blank line between each test case.

Output

For each test case,output the number of the candidate who wins the election, or -1 if the election ends without a winner.

Sample Input

5
120
102
210
021
012

8
3120
3012
1032
3120
2031
2103
1230
1230

Sample Output

1
-1

Hint:
Case 1:
Nobody gets an absolute majority in the first round and candidate 2 is eliminated. Candidate 1 then receives 3 votes in the next round, giving an absolute majority.

Case 2:
Candidate 0 is eliminated in the first round of voting. Candidate 2 is eliminated in the second round. In the third round, candidates 1 and 3 get 4 votes each. Neither candidate receives an absolute majority, and they are both eliminated for having the least number of votes, so the election ends without a winner.

�q�题的投��，每轮都是从第一个数开始找的，扑ֈ��W�一个没有被淘汰的�h�?br>我一开始以��Z��前几轮选的人，在后面不能被选了。。。�?

chinaeli 2008-03-31 20:57 发表评论

chinaeli — Thu, 27 Mar 2008 09:24:00 GMT

#include<iostream>
#include<string>
#include<iomanip>
#include<vector>
#include<algorithm>
const int MAXN=100;
using namespace std;
int cmp ( const void* p1 , const void *p2 )
{
    return *( double* )p1>*(double*)p2?1:-1;
}
typedef struct COLOR
{
    string col;
    double mean,median,sum,value[MAXN];
    int num,p;
}COL;
int find ( vector<COL> &str , COL s )
{
    for ( vector<COL>::size_type i = 0; i != str.size(); i++ )
        if ( str[i].col==s.col )
        {
            str[i].sum+=s.sum;
            str[i].num++;
            str[i].value[str[i].p++]=s.sum;
            return 1;
        }
    return 0;
}
void add ( vector<COL>& str , COL s )
{
    if ( !find ( str , s ) )
    {
        s.p=1;
        s.num=1;
        s.value[0]=s.sum;
        str.push_back(s);
    }
}
void output ( vector<COL> str )
{
    double s=0,v[MAXN],median;
    int n=0,q,j=0;
    for (vector<COL>::size_type i = 0; i != str.size(); ++i )
    {
        for (q=0 ; q<str[i].p ; q++)
            v[j++]=str[i].value[q];
        s+=str[i].sum;
        n+=str[i].p;
        qsort(str[i].value,str[i].num,sizeof(str[i].value[0]),cmp);
        if (str[i].num%2)
            str[i].median=str[i].value[(str[i].num-1)/2];
        else
            str[i].median=(str[i].value[str[i].num/2]+str[i].value[str[i].num/2-1])/2;
        cout<<str[i].col<<"\t"<<": "<<"sum="<<"\t"
            <<setw(10)<<str[i].sum<<"\t"<<"mean="<<"\t"
            <<setw(10)<<str[i].sum/str[i].num<<"\t"<<"median="<<"\t"
            <<setw(10)<<str[i].median<<endl;

        /*
        for ( int j=0 ; j<str[i].p ; j++ )
            cout<<str[i].value[j]<<endl;
            */
    }
    qsort(v,j,sizeof(v[0]),cmp);
    if (n%2!=0)
        median=v[(n-1)/2];
    else
        median=(v[n/2]+v[n/2-1])/2;

    cout<<"============================================================================"<<endl;
    cout<<"ALL"<<"\t"<<": "<<"sum="<<"\t"
        <<setw(10)<<s<<"\t"<<"mean="<<"\t"
        <<setw(10)<<s/n<<"\t"<<"median="<<"\t"
        <<setw(10)<<median<<endl;
    /*
    for ( j=0 ; j<n ;j++ )
        cout<<v[j]<<endl;
        */
}
int main ( )
{
    vector<COL> str;
    COL s;
    while ( cin>>s.col>>s.sum )
    {
        add(str,s);
    }
    output( str );
}

chinaeli 2008-03-27 17:24 发表评论

chinaeli — Thu, 27 Mar 2008 04:35:00 GMT

上面�q�个代码首先通过main函数的参�?/span>*args[]��d��key�Q�利�?/span>(i++)%len实现�?/span>key各个字符的��@环操作。利�?/span>cin.get( )依次��d��各个字符�Q��ƈ�?/span>key中的字符�q�行异或�q�算�Q�得到加密的字符。由于异或运��的可逆性，�?/span> a==(a^b)^b ,可以恢复得到原文�?span style="FONT-SIZE: 12pt; FONT-FAMILY: 宋体; mso-ascii-font-family: 'Times New Roman'; mso-hansi-font-family: 'Times New Roman'; mso-bidi-font-family: 'Times New Roman'; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">在调试过�E�中发现�Q�加密会得到一�?/span>ASCII码编号�ؓ26的字�W�，然后在解密的时候，若读入这个编码�ؓ26的字�W�，�E�序��׃��l�止。所以第一个想法就是忽略所有非打印字符�Q�在�~�码�Ӟ��若所得密码�ؓ非打印字�W�，则不�q�行加密�Q�输出原文。但是受��C��老师的否认。于是再作修改，�l�实验发玎ͼ�只有�~�码�?/span>26的字�W�会出现�q�种奇怪的情况�Q�因此决定将�q�个字符定义成编码�ؓ27的字�W�，�?/span>’u’^’o’会��?/span>26的字�W�，��把�q�个字符�Ҏ��27�Q�结果最后出现原来应该是’o’的，解密得到�?/span>’n’�Q�但是没有其他更好的办法了，��M��有点误差的嘛�?br>
以下��Z��改后的代�?br>

#include<iostream>
#include<cstring>
using namespace std;
const int MAXN=100000;
int main ( int argc , char *args[] )
{
    char p,code;
    int i=0,l;
    char key[MAXN];
    if ( argc<2 )
        while ( cin.get(p)   )
            cout.put(p);
    else
    {
        strcpy(key,args[1]);
        l=strlen(key);
        while ( cin.get(p)  )
        {
            code=p^key[i%l];
            if ( code==26 )
                code=27;
            cout<<code;
            i++;
        }
    }
}

chinaeli 2008-03-27 12:35 发表评论

详细解说 STL 排序(Sort)

chinaeli — Wed, 26 Mar 2008 18:57:00 GMT

摘要: 0 前言: STL�Q��ؓ什么你必须掌握对于�E�序员来��_��数据�l�构是必修的一门课。从查找到排序，从链表到二叉树，几乎所有的��法和原理都需要理解，理解不了也要死记��背下来。幸�q�的是这些理论都已经比较成熟�Q�算法也基本固定下来�Q�不需要你再去��p��心思去考虑其算法原理，也不用再去验证其准确性。不�q�，�{�你开始应用计��机语言来工作的时候，你会发现�Q�面对不同的需求你需要一�ơ又一�ơ去用代码重复实现这些已�l�成熟的... 阅读全文

chinaeli 2008-03-27 02:57 发表评论

我可怜的�W�三�ơC++作业啊~只有70分~

chinaeli — Wed, 26 Mar 2008 17:06:00 GMT

助教�l�我的邮件中�q�样��_��

Jerry Huang �?�?
昄��详细信息 3�?1�?(6天前)

Hi,

我编译了你们的代码，�q�行了测试，好像和希望的�l�果差距比较大，请你们再��查确认一下�?/p>

如果是提交错了，请重新提交�?/p>

Thanks

huang

�l�果我只打了70分。。。。�?/p>

1#include<iostream>
2using namespace std;
3/* 判断字符串是否在引号里面 */
4int qutation ( char c , bool &f ,bool f1 )
5{
6    char temp;
7    if ( c=='"' && f1==false)
8    {
9        f=true;
10        cout<<c;
11        while ( f==true )
12        {
13            cin.get(temp);
14            if ( temp=='"')
15                f=false;
16            cout<<temp;
17        }
18        return 1; //发现引号
19    }
20    return 0; //没有发现引号
21}
22void cut_add  ( )
23{
24    char c,temp;
25    bool f1,f2,f3;
26    f1=f2=f3=false; // f1标记block注释,f2标记引号,f3标记line注释
27    while ( cin.get(c) )
28    {
29        if ( !qutation(c,f2,f1) )  // 没有出现引号
30        {
31            /*判断注释开�?/span>*/
32            if ( c=='/' )
33            {
34                cin.get(temp);
35                if ( temp=='*' )
36                    f1=true;//扑ֈ�了block注释的开�?br>37                else
38                    if ( temp=='/' )
39                        f3=true;//扑ֈ�了line注释的开�?br>40                /*当没有找到注释的开头时�Q�执行else部分*/
41                    else
42                    {
43                        cout<<c;
44                        cin.putback(temp);
45                    }
46            }
47            else
48                if ( f1==false && f3==false )
49                    cout<<c;
50            /*判断注释�l�尾*/
51            if ( c=='*' )
52            {
53                cin.get(temp);
54                if ( temp=='/' )
55                    f1=false; //关闭block注释
56                else
57                {
58                    cout<<c;
59                    cin.putback(temp);
60                }
61            }
62            else
63                if ( c=='\n' )
64                {
65                    f3=false; //关闭line注释
66                    cout<<c;
67                }
68        }
69    }
70}
71int main ( )
72{
73    cut_add();
74}

我不知道什么原因，手动输入可以的，但是用文件输入输出的话，输出��停不了了。。。�?br>

题目�Q?br>��是�l�你�?cpp文�g�Q�这是加注释的，然后让你生成一�?txt文�g�Q�除��M��码的注释�?br>用命令行输入�Q?br>erasecomment < DataIn.cpp > result.txt

chinaeli 2008-03-27 01:06 发表评论

欧美激情影音先锋,国外成人在线,性欧美videos另类喷潮

push!

11

详细解说 STL 排序(Sort)

我可怜的�W�三�ơC++作业啊~~~只有70分~~~

我可怜的�W�三�ơC++作业啊~只有70分~