http://acm.hdu.edu.cn/showproblem.php?pid=5154

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3073 Accepted Submission(s): 1183

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

Sample Output

YES
NO

Source

BestCoder Round #25

Recommend

heyang | We have carefully selected several similar problems for you: 6447 6446 6445 6444 6443

思路

题目大意：哈利用一个魔法电脑处理N个任务，但是有M个前后关系(a，b)，意思是在b执行之前必须先执行a，

即a任务在b任务前，问你是否能满足要求处理完这N个任务。（也就是判断扔进队列的数目是否为n?）

思路：拓扑排序。裸的模板题，需要理解拓扑排序实质的流程。

因为拓扑排序本身用到了队列。它先将所有入度为0的点入队，并用num统计入度不为0的点。

遍历队列中的点所连的所有边，并减少该点连接边另一端的入度，只要另一端入度为0了，就将它加入队列中，

最后比较n和num是否相等就可以判断是否能处理完这N个任务。

AC Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int nmax=1010;
vector<int>G[nmax];
int inDegree[nmax];
int n,m;

bool toposort(){
	int num=0;
	queue<int>q;
	while(!q.empty()) q.pop();
	for(int i=1;i<=n;i++){//顶点编号：1~N
		if(inDegree[i]==0){
			q.push(i);//把入度为0的顶点扔进队列 
		} 
	}
	while(!q.empty()){
		int u=q.front();
		q.pop();
		num++;
		for(int i=0;i<G[u].size();i++){
			int v=G[u][i];
			inDegree[v]--;
			if(inDegree[v]==0){
				q.push(v);
			}
		}
	}
	if(num==n) //没有环路 
		return true;
	else		//有环路 
		return false;
	
}


int main(int argc, char** argv) {
	while(scanf("%d%d",&n,&m)!=EOF){
		memset(inDegree,0,sizeof(inDegree));
		for(int i=1;i<=n;i++){
			G[i].clear();
		}
		int u,v;
		for(int i=0;i<m;i++){
			scanf("%d %d",&u,&v);
			G[u].push_back(v);
			inDegree[v]++;
		}
		if(toposort()) puts("YES");
		else puts("NO");
	} 
	return 0;
}

【拓扑排序专题】HDU 5154 Harry and Magical Computer