浙江财经大学
信息管理与工程学院

HDU 4324 Triangle LOVE(拓扑排序)

本文由 Ocrosoft 于 2016-07-21 17:26:05 发表

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4423    Accepted Submission(s): 1755

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

已经给出矩阵的形式了。

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <list>
#include <stack>
#include <string>
#include <vector>
#include <cstring>
#include <cctype>
#include <queue>
#include <set>
#include <climits>
#define ms(a) memset(a,0,sizeof(a))
using namespace std;
int n,t;
char mp[2001][2001];//邻接表保存边信息
int indegree[2001];//保存入度
queue<int> q;
bool toposort()
{
    int cnt=0;
    while(!q.empty())
    {
        t=q.front(),q.pop(),cnt++;
        for(int i=0; i<n; i++)
            if(mp[t][i]&&!(--indegree[i]))
                q.push(i);
    }
    if(cnt==n)return 1;
    return 0;
}
int main()
{
    int N;
    cin>>N;
    for(int I=1; I<=N; I++)
    {
        cin>>n;
        ms(indegree);
        for(int i=0; i<n; i++)
        {
            scanf("%s",mp[i]);
            for(int j=0;j<n;j++)
                if(mp[i][j]=='1')indegree[j]++;
        }
        while(!q.empty())q.pop();
        for(int i=0; i<n; i++)
            if(!indegree[i])q.push(i);
        bool ans=toposort();
        printf("Case #%d: %s\n",I,ans?"No":"Yes");
    }
    return 0;
}

 

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