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 23 12 13 33 22 11 3

Sample Output

YESNO

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;}