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=2010;
int map[MAXN][MAXN];
int que[MAXN],flot,N,tot=0;
char c[MAXN][MAXN];
queue<int>dl;
void topsort(){
    for(int i=1;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]=-1;
       // printf("%d",dl.size());
        for(int j=1;j<=N;j++){
            if(map[k][j])que[j]--;
            if(!que[j])dl.push(j);
        }
    }
   // puts("asfaf");
    if(flot==N)printf("Case #%d: No
",tot);
    else printf("Case #%d: Yes
",tot);
}
void initial(){
    memset(que,0,sizeof(que));
    while(!dl.empty()){
        dl.pop();
    }
    flot=0;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
            tot++;
            initial();
        scanf("%d",&N);
        //puts("asfaf");
        //printf("%d
",N);
        for(int i=0;i<N;i++)scanf("%s",c[i]);
        for(int i=1;i<=N;i++){
            for(int j=1;j<=N;j++){
            map[i][j]=c[i-1][j-1]-'0';
                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 0;
}