HDU - 4324 Triangle LOVE（拓扑排序 + 判断环路）

题意：人类间的凄美爱情，若a爱b，则b一定不爱a， 求是否存在三角恋，判断是否存在环路即可。

解题思路：

直接链式向前星存边，手工队列拓扑排序，判断环路即可。

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

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

using namespace std;

int buf[10];
//整型变量快速输入输出函数
inline int readint()
{
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c))
    {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

inline void writeint(int i)
{
    int p = 0;
    if(i == 0) p++;
    else while(i)
    {
        buf[p++] = i % 10;
        i /= 10;
    }
    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);
}
//////////////////////////////////////////////////////////////////////
#define MAX_N 2005
const int INF = 0x3f3f3f3f;
int n;
char m[MAX_N][MAX_N];
int indegree[MAX_N];
int head[MAX_N];
int top;
struct node
{
    int to, next;
}edge[MAX_N * MAX_N];

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

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

int topo_sort()
{
    int iq = 0;
    int que[MAX_N];
    for(int i = 0 ; i < n ; i++)
    {
        if(indegree[i] == 0)
        {
            que[iq++] = i;
        }
    }

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

}

int main()
{
    int t;
    t = readint();
    int cas = 1;
    while(t--)
    {
        init();
        n = readint();
        for(int i = 0 ; i < n ; i++)
        {
            scanf("%s", &m[i]);
            for(int j = 0 ; j < n ; j++)
            {
                if(m[i][j] == '1')
                {
                    add_edge(i, j);
                    indegree[j]++;

            }
        }
        int ans = topo_sort();
        if(ans) printf("Case #%d: Yes\n", cas++);
        else printf("Case #%d: No\n", cas++);
    }

    return 0;
}