HDOJ 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): 1822    Accepted Submission(s): 716

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

 

Source

BestCoder Round #25

 

    拓扑排序模板题,判断是否成环就可以了

#include<iostream>#include<cstring>#include<cstdio>#include<queue>#define MAX 505using namespace std;int mmap[MAX][MAX];int indegree[MAX];int TopSort(int n){    queue<int>q;    int i,j,t,Count=0;    for(i=1;i<=n;i++)        if(indegree[i]==0)        q.push(i);    while(!q.empty())    {        t=q.front();q.pop();Count++;        for(i=1;i<=n;i++)        {            if(mmap[t][i]){                    indegree[i]--;                    if(indegree[i]==0)                        q.push(i);                }        }    }    return Count;}int main(){    int n,m,a,b,i,j;    while(~scanf("%d%d",&n,&m))    {        memset(mmap,0,sizeof(mmap));        memset(indegree,0,sizeof(indegree));        for(i=1;i<=m;i++)        {            scanf("%d%d",&a,&b);            mmap[a][b]=1;        }        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)                if(mmap[i][j])                indegree[j]++;        printf("%s\n",TopSort(n)==n?"YES":"NO");    }    return 0;}