HDU NO.4324 Triangle LOVE（拓扑排序）

题意：

如果A爱B，那么B就一定不爱A --!.

大致意思是，在图中能不能找到三角环。

原题描述：

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

2
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110

Sample Output

Case #1: Yes
Case #2: No

思路：

可以吧图中是“1”的坐标构成有向图，然后拓扑排序，如果结果没有拓扑序列即有环也就是存在三角恋 --！

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<stack>
#include<cstring>
#include<string>
#include<vector>
#include<set>

using namespace std;

#define X first
#define Y second

const int INF = 0x3f3f3f3f;
const int MAX = 3000;
int head[MAX], indegree[MAX], que[MAX];
int top, n;

struct node{
int to, w, next;
}edge[MAX * MAX];

void add_edge(int u, int v)
{
edge[top].to = v;
edge[top].next = head[u];
head[u] = top++;
}

void init(){
top = 0;
memset(head, -1, sizeof(head));
memset(indegree, 0, sizeof(indegree));
}

bool Topsort(){
int iq = 0;
for(int i = 0; i < n; i++)
if(indegree[i] == 0)
que[iq++] = i;
for(int i = 0; i < iq; i++){
for(int j = head[que[i]]; j != -1; j = edge[j].next){
indegree[edge[j].to]--;
if(indegree[edge[j].to] == 0)
que[iq++] = edge[j].to;
}
}
if(iq < n - 1)
return true;
else
return false;
}

int main(){
int t, cns = 1;
scanf("%d", &t);
while(t--){
init();
scanf("%d", &n);
char arr

;
for(int i = 0; i < n; i++){
scanf("%s", &arr[i]);
for(int j = 0; j < n; j++){
if(arr[i][j] == '1'){
add_edge(i, j);
indegree[j]++;
}
}
}
int res = Topsort();
if(res)
printf("Case #%d: Yes\n", cns++);
else
printf("Case #%d: No\n", cns++);
}
return 0;
}