杭电4324--Triangle LOVE（拓扑排序）

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3479    Accepted Submission(s): 1350

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). 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

 

 

Author

BJTU

 

 

Source

2012 Multi-University Training Contest 3

 

 

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RE:三角恋 → → 判断是否成环。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
char map[2020][2020]; 
int indegree[2020];
int n; 
bool Solve()
{
    queue<int> q;
    for(int i = 0; i < n; i++)
    {
        if(indegree[i] == 1)
            q.push(i);
    }
    int num = 0;
    while(!q.empty())
    {
        int temp = q.front();
        q.pop();
        num++;
        for(int i = 0; i < n; i++)
        {
            if(map[temp][i])
            {
                indegree[i]--;
                if(indegree[i] == 0)    
                q.push(i);
            }
        } 
    }
    if(num == n)
        return false;
    else
        return true;
} 
int main()
{
    int t, temp = 1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        memset(indegree, 0, sizeof(indegree));
        for(int i = 0; i < n; i++)
        {
            scanf("%s", map[i]);
            for(int j = 0; j < n; j++)
                if(map[i][j] == '1')
                    indegree[j]++;
        }
        if(Solve())
            printf("Case #%d: Yes
", temp++);
        else
            printf("Case #%d: No
", temp++);
    } 
    return 0;    
}  

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
char map[2020][2020], indegree[2020];
int n;
bool Tsort()
{
    int m;
    for(int i = 0; i < n; i++)
    {
        m = -1;
        for(int j = 0; j < n; j++)
        {
            if(indegree[j] == 0)
            {
                m = j;
                break;
            }
        }
        if(m == -1)
            return true;
        indegree[m]--;
        for(int j = 0; j < n; j++)
        {
            if(map[m][j] == '1')
                indegree[j]--;
        }
    }
    return false;
}
int main()
{
    int t, Q = 1;
    scanf("%d", &t);
    while(t--)
    {
        memset(indegree, 0, sizeof(indegree));
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%s", map[i]);
            for(int j = 0; j < n; j++)
                if(map[i][j] == '1')
                    indegree[j]++;    
        }    
        if(Tsort())
            printf("Case #%d: Yes
", Q++);
        else
            printf("Case #%d: No
", Q++);
    }
    return 0;
}