HDU 4324 Triangle LOVE 拓扑排序

题目描述：

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 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

25001001000001001111011100050111100000010000110001110 

Sample Output

Case #1: YesCase #2: No 

题目大意：

这道题是个悲伤的故事。
给定n个人，n*n的关系矩阵，a[i][j]为1表示i喜欢j，0表示不喜欢。没有自恋和情投意合的情况。求是否有三角恋。（不相信真爱了）

这道题其实就是拓扑排序求是否有环。我用的是遍历各个点的入度（喜欢他的人），如果每个点的入度都大于0，必然有环。将入读为0的点以及边删除，继续寻找，如果点全部删除了还没有，那就没有环了。。

代码如下：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;typedef long long ll;const int INF = 0x3f3f3f3f;const double eps = 1e-4;const int MAXN=2020;char mp[MAXN][MAXN];int in[MAXN];int T;int n;bool topsort(){    int i,j,k;    for(i=1; i<=n; i++)    {        for(j=1; j<=n; j++)        {            if (in[j]==0) break;//入度为0的点        }        if (j==n+1) return true;//寻找完毕        else        {            in[j]--;//删点            for(k=1; k<=n; k++)            {                if (mp[j][k]=='1' && in[k]!=0) in[k]--;                //删边            }        }    }    return false;}int main(){    scanf("%d",&T);    for(int t=1; t<=T; t++)    {        memset(in,0,sizeof(in));        scanf("%d",&n);        for(int i=1; i<=n; i++)        {            scanf("%s",mp[i]+1);            for(int j=1; j<=n; j++)            {                if (mp[i][j]=='1')  in[j]++;            }        }        if (topsort()==true) printf("Case #%d: Yes\n",t);        else printf("Case #%d: No\n",t);    }    return 0;}