题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4324
Triangle LOVE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3858 Accepted Submission(s): 1516
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). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,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
Author
BJTU
Source
2012 Multi-University Training Contest 3
题目大意：给一个n*n的矩阵，Map[i][j]表示i喜欢j；在这个矩阵表示的范围内，找到是否存在三角恋的关系。
解题思路：做的第一道拓扑排序判环题目。判断其中是否有环就可以了，如果存在环必然存在三元环（这个是根据这道题目得出的结论）。
先介绍一下拓扑排序：对一个有向无环图(Directed Acyclic Graph简称DAG)G进行拓扑排序，是将G中所有顶点排成一个线性序列，使得图中任意一对顶点u和v，若边(u,v)∈E(G)，则u在线性序列中出现在v之前。简单的说，由某个集合上的一个偏序得到该集合上的一个全序，这个操作称之为拓扑排序。
再通俗一点就是，一个点一个点的去删除，删掉的必须是入度为0的，如果将入度为0的全部删完，还有剩余边的话就会产生死锁，其中必存在环。
详见代码。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4
5 using namespace std;
6
7 int indir[2010];//用来表示入度
8 char Map[2010][2010];//存储的是i,j两个节点的关系，1:i love j,0:j love i
9
10 int main()
11 {
12 int t;
13 scanf("%d",&t);
14 int ff=1;
15 while (t--)
16 {
17 memset(indir,0,sizeof(indir));
18 int flag=0;
19 int n;
20 scanf("%d",&n);
21 for (int i=0; i<n; i++)
22 {
23 scanf("%s",Map[i]);
24 for (int j=0; j<n; j++)
25 {
26 if (Map[i][j]==‘1‘)//i love j,也就是i指向j
27 indir[j]++;//j的入度++
28 }
29 }
30 int j;
31 for (int i=0; i<n; i++)
32 {
33 for (j=0; j<n; j++)
34 if (indir[j]==0)
35 break;
36 if (j==n)//如果全部找完都没有发现入度为0的必存在环。
37 {
38 flag=1;
39 break;
40 }
41 else
42 {
43 indir[j]--;//j的入度删掉
44 for (int k=0;k<n;k++)//j指向的点删掉
45 {
46 if (Map[j][k]==‘1‘)
47 {
48 indir[k]--;
49 }
50 }
51 }
52 }
53 if (flag==1)
54 printf ("Case #%d: Yes\n",ff++);
55 else
56 printf ("Case #%d: No\n",ff++);
57 }
58 }