poj3687

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9028		Accepted: 2444

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 toN in such a way that:

1. No two balls share the same label.
2. The labeling satisfies several constrains like "The ball labeled with a is lighter than the one labeled withb".

Can you help windy to find a solution?

Input

The first line of input is the number of test case. The first line of each test case contains two integers,N (1 ≤N ≤ 200) and M (0 ≤ M ≤ 40,000). The nextM line each contain two integersa and b indicating the ball labeled witha must be lighter than the one labeled withb. (1 ≤ a, b ≤N) There is a blank line before each test case.

Output

For each test case output on a single line the balls' weights from label 1 to labelN. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on... If no solution exists, output -1 instead.

Sample Input

5

4 0

4 1
1 1

4 2
1 2
2 1

4 1
2 1

4 1
3 2

Sample Output

1 2 3 4
-1
-1
2 1 3 4
1 3 2 4

#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
typedef struct v{

	int vex;
	v *next;

}V;
V*p;
typedef struct h{

	int indegree;
	v *next ;
}H;
int n,result[300],tot,ans[300];
priority_queue<int ,vector<int>,less<int> > q;
H team[50000];
bool visit[305][305]; 
void topsort()
{
	int i,ci,sum,a,b;
	ci=1;
	for(i=1;i<=n;i++)
	{
	
		if(team[i].indegree==0)
			q.push(i);
	
	}
	sum=n;
	
	while(!q.empty())
	{
		sum--;
		a=q.top();
		q.pop();
		result[ci++]=a;
		for(p=team[a].next;p!=0;p=p->next)
		{
		
			b=p->vex;
			if(--team[b].indegree==0)
				q.push(b);
		}

	
	}
//	printf("%d\n",sum);
	if(sum>0)
	{
			printf("-1\n");
	}
	else
	{
		int nn=n;
		for(i=1;i<=n;i++)
		{
		
			ans[result[i]]=nn--;//这是排序
		}
	for(i=1;i<n;i++)
	
			printf("%d ",ans[i]);//这是重量
		printf("%d\n",ans[i]);
	
	}

}
int main()
{

	int i,m,a,b,T,j;

	scanf("%d",&T);
	while(T--)
	{
		while(!q.empty())
		{
		
			q.pop();
		}

		scanf("%d%d",&n,&m);
		for(i=0;i<=n;i++)
		for(j=0;j<=n;j++)
		{
		
			visit[i][j]=false;
		}
	
		for(i=0;i<=n;i++)
		{
			team[i].indegree=0;
			team[i].next=NULL;
			result[i]=0;
		}
		while(m--)
		{
		
			scanf("%d%d",&b,&a);//反向建表
			if(visit[a][b])
			continue;
			visit[a][b]=true;
			team[b].indegree++;
			p=new V;
			p->vex=b;
			p->next=team[a].next;
			team[a].next=p;

		}

		topsort();

	}


	return 0;
}