HDOJ 5154 Harry and Magical Computer floyd判环

 floyd判环

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1005    Accepted Submission(s): 404

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

 

/* ***********************************************Author        :CKbossCreated Time  :2015年02月15日 星期日 22时23分31秒File Name     :HDOJ5154.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;const int maxn=110;const int INF=0x3f3f3f3f;int n,m;int g[maxn][maxn];int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);while(scanf("%d%d",&n,&m)!=EOF){memset(g,63,sizeof(g));for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);g[a][b]=1;}for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)g[i][j]=min(g[i][j],g[i][k]+g[k][j]);bool flag=false;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(g[i][j]==g[j][i]&&g[i][j]!=INF){flag=true;break;}}if(flag) break;}if(flag) puts("NO");else puts("YES");}        return 0;}