hdu 5154(拓扑排序)

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2017    Accepted Submission(s): 801

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

 

题意:有n个进程,n个进程有m条联系,如果(a,b)那么代表b要在a之前完成,现有m条联系,问所有的进程能否都完成??

题解:这个题开始想简单了,以为只要判环就行了,用并查集去做果然WA，然后发现这是有向边，所以我们可以采用拓扑排序,看最后拓扑排序进入队列的点是否为n，如果是,那么都可以完成,如果不是,那么就有一些陷入了死循环。

给一组数据:

3 3
3 1
2 1
3 2

ans:YES

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
#include <queue>
using namespace std;
int indegree[105];
struct Edge{
    int v,next;
}edge[10005];
int head[105];
int tot;
void addEdge(int u,int v,int &k){
    edge[k].v = v,edge[k].next = head[u],head[u] = k++;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        memset(indegree,0,sizeof(indegree));
        memset(head,-1,sizeof(head));
        tot=0;
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            addEdge(v,u,tot);
            indegree[u]++;
        }
        queue<int >q;
        for(int i=1;i<=n;i++){
            if(indegree[i]==0) q.push(i);
        }
        int cnt = 0;
        while(!q.empty()){
            int u = q.front();
            cnt++;
            q.pop();
            for(int k=head[u];k!=-1;k=edge[k].next){
                indegree[edge[k].v]--;
                if(!indegree[edge[k].v]) q.push(edge[k].v);
            }
        }
        if(cnt==n) printf("YES
");
        else printf("NO
");
    }
    return 0;
}
/**
3 3
3 1
2 1
3 2
*/