Triangle LOVE （hdu4324）-CFANZ编程社区

题目链接：Triangle LOVE （http://acm.hdu.edu.cn/showproblem.php?pid=4324）

题目描述：

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.

大意：最近，科学家发现两个人之间有爱。例如，在A和B之间，如果A不爱B，那么B必须爱A，反之亦然。两个人相爱是不可能的，这是一个多么疯狂的世界！
现在，科学家们想知道N人之间是否存在“三角恋”。三角恋爱是指任何三个人（A，B和C），A爱B，B爱C和C爱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

Triangle LOVE （hdu4324）_i++

一道英文题，题目意思还算好理解，就是判断给定的图有没有形成环，若形成环则为三角恋（多角恋也是，不过题目应该不会给出多角恋的情况），正好刚复习了拓扑排序，进行拓扑排序，若可以完成排序，则证明是无环的，即没有三角恋。

若是有环，则为三角恋。

但是这道题却坑了我好久，能有一个多小时了吧....在后面归纳下进去的那些坑。

先贴上交了不下10遍终于AC的代码：

#include <stdio.h>
#include <memory.h>
#include<stdbool.h>
#define MAXV 2010
char map[MAXV][MAXV];
int indegree[MAXV];
int cnt=0;
void topo_sort(int n)
{
    int i,j,k;
    for(i=0; i<n; i++)
    {
        for(j=0; j<n; j++)
        {
            if(indegree[j]==0)
            {
                break;
            }
        }
        if(j==n)
        {
            return;
        }
        else
        {
            indegree[j]--;
            cnt++;
            for(k=0; k<n; k++)
            {
                if(map[j][k]=='1'&&indegree[k]!=0)
                    indegree[k]--;
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    int q;
    for(q=1;q<=t;q++)
    {
        cnt=0;
        memset(map,0,sizeof(map));
        memset(indegree,0,sizeof(indegree));
        int n;
        scanf("%d",&n);
        int i,j;
        for(i=0; i<n; i++)
        {
            scanf("%s",map[i]);
            for(j=0; j<n; j++)
                if(map[i][j]=='1')
                    indegree[j]++;
        }
        topo_sort(n);
        if(cnt!=n)
            printf("Case #%d: Yes\n",q);
        else
            printf("Case #%d: No\n",q);
    }
    return 0;
}

一开始用数组储存的邻接矩阵，但总是

Time Limit Exceeded。

于是换成char型数组进行存储，可还是存在着一些问题。

for(j=0; j<n; j++)
        {
            if(indegree[j]==0)
            {
                break;
            }
        }
        if(j==n)
        {
            return;
        }