Legal or Not
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11382 Accepted Submission(s): 5346
We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless,some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.
Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.
If it is legal, output "YES", otherwise "NO".
0 1
1 2
2 2
0 1
1 0
0 0
NO
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[]= {,,,,,,,,,,,,};
int mon2[]= {,,,,,,,,,,,,};
int dir[][]= {{,},{,-},{,},{-,}}; int getval()
{
int ret();
char c;
while((c=getchar())==' '||c=='\n'||c=='\r');
ret=c-'';
while((c=getchar())!=' '&&c!='\n'&&c!='\r')
ret=ret*+c-'';
return ret;
} #define max_v 105
int indgree[max_v];
vector<int> vv[max_v];
int n,m;
queue<int> q;
int tpsort()
{
while(!q.empty())
q.pop();
for(int i=;i<=n;i++)
if(indgree[i]==)
q.push(i);
int c=;
int temp;
while(!q.empty())
{
temp=q.front();
q.pop();
c++;
for(int i=;i<vv[temp].size();i++)
{
indgree[vv[temp][i]]--;
if(indgree[vv[temp][i]]==)
q.push(vv[temp][i]);
}
}
if(c!=n)//判环 拓扑完之后,如果存在点没有入队,那么这个点一定是环上的
return ;
else
return ;
}
int main()
{
/*
有向图判环 拓扑排序
无向图判环 并查集
*/
int x,y;
while(~scanf("%d %d",&n,&m))
{
if(n==&&m==)
break;
memset(indgree,,sizeof(indgree));
for(int i=;i<=n;i++)
vv[i].clear();
int flag=;
for(int i=;i<=m;i++)
{
scanf("%d %d",&x,&y);
x++,y++;
if(x==y)
continue;
if(count(vv[x].begin(),vv[x].end(),y)==)//防重边
{
vv[x].push_back(y);
indgree[y]++;
}
if(count(vv[y].begin(),vv[y].end(),x)!=)//环的一种
{
flag=;
}
}
flag=tpsort();
if(flag)
printf("NO\n");
else
printf("YES\n");
}
return ;
}