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;

}