Legal or Not HDU - 3342 (拓扑排序)

 

 

  注意点： 输入数据中可能有重复，需要进行处理！

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <sstream>


using namespace std;

int n, m;

int indegree[101];
int g[101][101]; 

int main()
{
    while(scanf("%d %d", &n, &m) != EOF)
    {
        if(n == 0 && m == 0)
            break;
        
        memset(indegree, 0, sizeof(indegree));
        memset(g, 0, sizeof(g));
        
        int a, b;
        for(int i = 1; i <= m; ++i)
        {
            scanf("%d %d", &a, &b);
            
            if(g[a][b] == 0)    // 输入中有可能有反复输入如1,2；1,2这种情况！！！ 
            {
                g[a][b] = 1;
                indegree[b]++;
            }
            
        }
        
        int cnt = 0;
        
        for(int k = 0; k < n; ++k)
        {
            int i, j; 
            for(i = 0; i < n; ++i)
            {
                if(indegree[i] == 0)
                {
                    indegree[i] = -1;
                    cnt++;
                    break;
                }
            } 
            
            for(j = 0; j < n; ++j)
            {
                if(g[i][j] == 1)
                {
                    indegree[j]--;
                    g[i][j] == 0;
                }
            }
        }
        
        
        if(cnt == n)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
       
    return 0;
}