HDOJ 5154 Harry and Magical Computer floyd判环 Harry and Magical Computer

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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

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Sample Output

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1
2
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4

YES
NO

Source

BestCoder Round #25

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/* ***********************************************
Author
:CKboss
Created 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;
}