Triangle LOVE 【topo 判有无环】

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 i,j = 1 means i-th people loves j-th people, otherwise A i,j = 0.
It is guaranteed that the given relationship is a tournament, that is, A i,i= 0, A i,j ≠ A j,i(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

题意 给一个有向图，问其中是不是有环。
可以用topo
（无语的是，我用链式向前星存图，却TLE改为邻接表存，就能过。）
代码

#include<stdio.h>#include<string.h>#include<queue>using namespace std;const int MAXN =2000+100;vector<int>G[MAXN];int n;int in[MAXN];void init(){    for(int i=0;i<=n;i++)    {        G[i].clear();        in[i]=0;    }}void getmap(){    char s[MAXN];    for(int i=1;i<=n;i++){        scanf("%s",s);        for(int j=0;j<n;j++){            if(s[j]=='1') {                G[i].push_back(j+1);                in[j+1]++;            }        }    }}bool topo(){    queue<int>Q;    for(int i=1;i<=n;i++)        if(in[i]==0) Q.push(i);        int num=0;    while(!Q.empty()){        int now=Q.front();Q.pop();num++;        for(int i=0;i<G[now].size();i++){            if(--in[G[now][i]]==0)                Q.push(G[now][i]);        }    }    return num!=n;}int main(){    int k=1;    int t;scanf("%d",&t);while(t--){        scanf("%d",&n);        init();        getmap();        printf("Case #%d: %s\n",k++,topo()?"Yes":"No");    }}