拓扑排序 杭电5154 Harry and Magical Computer

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 265    Accepted Submission(s): 123

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
  
    

 

#include <iostream>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
using namespace std;
#define lson rt<<1,l,MID
#define rson rt<<1|1,MID+1,r
//#define lson root<<1
//#define rson root<<1|1
#define MID ((l+r)>>1)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=10005;
const int base=1000;
const int inf=999999;
const double eps=1e-5;

int d[maxn];
vector<int> G[maxn];
bool tupo_sort(int n)
{
    stack<int> s;
    for(int i=1;i<=n;i++)
        if(d[i]==0)s.push(i);
    int cnt=0;
    while(!s.empty())
    {
        int m=s.top();//printf("%d ",m);
        s.pop();
        cnt++;
        for(int i=0;i<G[m].size();i++)
        {
            d[G[m][i]]--;
            if(d[G[m][i]]==0)s.push(G[m][i]);
        }
    }
    if(cnt<n)return false;
    return true;
}


int main()
{
    int n,m,i,j,k,t;
    while(~scanf("%d%d",&n,&m))
    {

        memset(d,0,sizeof(d));
        for(i=0;i<=n;i++)//每次使用注意 清空  这里忘记了 错误了很多次
            G[i].clear();
        while(m--)
        {
            int s,e;
            scanf("%d%d",&s,&e);
            G[e].push_back(s);
            d[s]++;
        }
    
        if(tupo_sort(n))
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}