HDU 5154 Harry and Magical Computer 拓扑排序 - c++编程基础

水题不解释

拓扑排序判断有无环

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. $1 \leq n \leq 100, 1 \leq m \leq 10000$
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). $1 \leq a, b \leq n$

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

#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        #include
       
         #include
        
          using namespace std; struct node { int u,v,next; } edge[1100000]; int head[1000000],vis[200000],rudu[200000],cnt,n; void add(int u,int v) { edge[cnt].v=v; edge[cnt].next=head[u]; head[u]=cnt++; } void topu() { memset(vis,0,sizeof(vis)); int s=0; queue
         
          q; while(!q.empty()) q.pop(); for(int i=1; i<=n; i++) { if(rudu[i]==0) { q.push(i); vis[i]=1; } } while(!q.empty()) { int u=q.front(); q.pop(); s++; for(int i=head[u]; i!=-1; i=edge[i].next) { if(rudu[edge[i].v]) { rudu[edge[i].v]--; } if(rudu[edge[i].v]==0&&vis[edge[i].v]==0) { q.push(edge[i].v); vis[edge[i].v]=1; } } } if(s==n) printf("YES\n"); else printf("NO\n"); } int main() { int m; while(~scanf("%d %d",&n,&m)) { cnt=0; memset(rudu,0,sizeof(rudu)); memset(head,-1,sizeof(head)); for(int i=0; i