Triangle LOVE(拓扑排序)

Triangle LOVE

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 49   Accepted Submission(s) : 30

Problem Description

Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
  Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.

 

Input

The first line contains a single integer t (1 <= t <= 15), the number of test cases. For each case, the first line contains one integer N (0 < N <= 2000). In the next N lines contain the adjacency matrix A of the relationship (without spaces). A[sub]i,j[/sub] = 1 means i-th people loves j-th people, otherwise A[sub]i,j[/sub] = 0. It is guaranteed that the given relationship is a tournament, that is, A[sub]i,i[/sub]= 0, A[sub]i,j[/sub] ≠ A[sub]j,i[/sub](1<=i, j<=n,i≠j).

 

Output

For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”. Take the sample output for more details.

 

Sample Input

2 5 00100 10000 01001 11101 11000 5 01111 00000 01000 01100 01110

 

Sample Output

Case #1: Yes Case #2: No

代码：

 

 #include<stdio.h>
 #include<string.h>
 #include<queue>
 using namespace std;
 const int MAXN=;
 int map[MAXN][MAXN];
 int que[MAXN],flot,N,tot=;
 char c[MAXN][MAXN];
 queue<int>dl;
 void topsort(){
     for(int i=;i<=N;i++)
         if(!que[i])dl.push(i);
         //printf("%d",dl.size());
         //puts("asfaf");
     while(!dl.empty()){
         int k=dl.front();
         flot++;
         dl.pop();
         que[k]=-;
        // printf("%d",dl.size());
         for(int j=;j<=N;j++){
             if(map[k][j])que[j]--;
             if(!que[j])dl.push(j);
         }
     }
    // puts("asfaf");
     if(flot==N)printf("Case #%d: No\n",tot);
     else printf("Case #%d: Yes\n",tot);
 }
 void initial(){
     memset(que,,sizeof(que));
     while(!dl.empty()){
         dl.pop();
     }
     flot=;
 }
 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
             tot++;
             initial();
         scanf("%d",&N);
         //puts("asfaf");
         //printf("%d\n",N);
         for(int i=;i<N;i++)scanf("%s",c[i]);
         for(int i=;i<=N;i++){
             for(int j=;j<=N;j++){
             map[i][j]=c[i-][j-]-'';
                 if(map[i][j])que[j]++;
             }
         }
        // puts("asfaf");
        /* for(int i=1;i<=N;i++){
             for(int j=1;j<=N;j++){
             printf("%d",map[i][j]);
             }
                 puts("");
         }*/
         //puts("asfaf");
         topsort();
     }
     return ;
 }