hdu 5154 Harry and Magical Computer（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): 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

3 23 12 13 33 22 11 3

 

Sample Output

YESNO

 

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 1000int 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;}