拓扑排序判断

拓扑排序的实现方法：

                                     (1)从有向图中选择一个没有前驱(即入度为0)的顶点并且输出它.

                                     (2)从网中删去该顶点,并且删去从该顶点发出的全部有向边.

                                     (3)重复上述两步,直到剩余的网中不再存在没有前趋的顶点为止.

Harry and Magical Computer

 

 Accepts: 408

 

 Submissions: 1622

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 32768/32768 K (Java/Others)

Problem Description

In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

Input

There are several test cases, you should process to the end of file. For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤10000 The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n

Output

Output one line for each test case. If the computer can finish all the process print "YES" (Without quotes). Else print "NO" (Without quotes).

Sample Input

3 2
3 1
2 1
3 3
3 2
2 1
1 3

Sample Output

YES
NO

分析：判断是否可以拓扑排序，有向图中是否有环，最简单的拓扑排序算法。

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<stack>
using namespace std;
int arc[501][501] ;
int top ;
int used[501] ; // 标记访问过
int index[501] ; // 统计入度数组
int m , n ;
stack<int>st ;

bool istopsort(){
    int countt = 0 ;
    for(int i =1 ; i<= n ; i++){
        if(!index[i])
            st.push(i);
    }
    while(!st.empty()){
        int tt = st.top() ;
        index[tt] -- ;
        st.pop() ;
         countt ++ ;
         for(int i = 1 ; i<=n ;i++)
            if(arc[tt][i]){
               index[i]-- ;
               if(!index[i])
                 st.push(i) ;
            }
    }
    if(countt == n)
        return 1 ;
    else return 0 ;
}
int main()
{
    int a,b;
    while(scanf("%d%d", &n , &m) != EOF){
        memset(arc, 0, sizeof(arc)) ;
        memset(index , 0 , sizeof(index)) ;
        for(int i =1 ; i<=m ; i++){
            scanf("%d%d" , &a , &b) ;
            if(!arc[a][b]){
                arc[a][b] = 1 ;
                index[b] ++ ;}
        }

        if(istopsort())
            cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}