Harry and Magical Computer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 499 Accepted Submission(s): 233
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.
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
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).
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
3 1
2 1
3 3
3 2
2 1
1 3
Sample Output
YES
NO
NO
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100001
const int inf=0x7fffffff; //无限大
int map[][];
int vis[];
int flag[];
int main()
{
int n,m;
while(cin>>n>>m)
{
memset(map,,sizeof(map));
memset(vis,,sizeof(vis));
memset(flag,,sizeof(flag));
int a,b;
for(int i=;i<m;i++)
{
cin>>a>>b;
map[b-][a-]=;
flag[a-]=;
}
queue<int> q;
for(int i=;i<n;i++)
{
if(flag[i]==)
{
q.push(i);
vis[i]=;
}
}
int now;
int next;
while(!q.empty())
{
now=q.front();
for(int i=;i<n;i++)
{
if(map[now][i]==)
{
if(vis[i]==)
continue;
q.push(i);
vis[i]=;
}
}
q.pop();
}
int flag1=;
for(int i=;i<n;i++)
{
if(vis[i]==)
{
flag1=;
break;
}
}
if(flag1==)
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
}
}