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 23 12 13 33 22 11 3

 

Sample Output

YESNO

 

#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;}