hdu 5154 Harry and Magical Computer（拓扑排序）

最新推荐文章于 2023-05-02 11:11:33 发布

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本文介绍了一个关于魔法计算机处理任务的问题，计算机需要处理一系列任务，并且这些任务之间存在依赖关系。文章通过拓扑排序的方法判断了计算机是否能够顺利完成所有任务。

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2018 Accepted Submission(s): 802

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

题意：计算机有很多个任务，任务之间有依赖关系，即完成a任务必须先完成b任务，问你计算机能否完成全部的任务

思路：跑一遍拓扑排序，看是否能遍历到所有的点即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define N 1000
int d[N],vis[N];
struct Edge
{
    int u,v,next;
} edge[N*10];
int cnt,head[N];
void addedge(int u,int v)
{
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
void init()
{
    cnt=0;
    memset(head,-1,sizeof(head));
    memset(d,0,sizeof(d));
    memset(vis,0,sizeof(vis));
}
int main()
{
    int n,m;
    int u,v;
    while(~scanf("%d %d",&n,&m))
    {
        init();
        priority_queue<int> q;
        for(int i=0; i<m; i++)
        {
            scanf("%d %d",&u,&v);
            d[u]++;
            addedge(v,u);
        }
        for(int i=1; i<=n; i++)
            if(!d[i])
            {
                q.push(i);
                vis[i]=1;
            }
        while(!q.empty())
        {
            int u=q.top();
            q.pop();
            for(int i=head[u]; i!=-1; i=edge[i].next)
            {
                int v=edge[i].v;
                d[v]--;
                if(!d[v])
                    {
                        q.push(v);
                        vis[v]=1;
                    }
            }
        }
        int flag=0;
        for(int i=1; i<=n; i++)
                if(!vis[i])
                flag=1;
        if(flag) printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}