poj 3687 Labeling Balls【反向拓扑】

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12246		Accepted: 3508

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N 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 with b".

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 next M line each contain two integers a and b indicating the ball labeled with a must be lighter than the one labeled with b. (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 label N. 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

这道题搜了下题解，大神的思路就是清新脱俗;附上自己的理解分析
There is a blank line before each test case.  这句话可把我坑死了  还以为每组测试数据前都要打印出一行空格，pe了好久
题意：输入n，m代表有n个数参加排序，接下来m行输入数据x ，y表示x要在y的前边，但是重量小的（即数字小的）要尽量放到前边；
题解：此题要用反向拓扑，正向是肯定不行的，如数据
     5 3
     1 4
     4 2
     3 5
     正向的话输出结果为 3 1 5 2 4 显然不对，正确结果应该为 1 3 4 2 5 同时采用优先队列，重的最先输出，我们先将重的，赋给
     最先出队的（因为设置的优先队列是大的先出队）这样就可以把轻的尽量留在最后（因为是要逆序，所以留到后边，输出时是在前边）

#include<stdio.h>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;
int n,m;
int map[300][300];
int vis[300],a[300];
void getmap()
{
	int i,j;
	memset(vis,0,sizeof(vis));
	memset(map,0,sizeof(map));
	while(m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		if(!map[b][a])
		{
			map[b][a]=1;
		    vis[a]++;
		}
	}
}
void tuopu()
{
	int i,j,k,g=n;
	k=0;
	memset(a,0,sizeof(a));
	priority_queue<int>q;
	while(!q.empty())
	    q.pop();
	for(i=1;i<=n;i++)
	    if(!vis[i])
	        q.push(i);
	int u;
	while(!q.empty())
	{
		u=q.top();
		q.pop();
		a[u]=g--;
		for(i=1;i<=n;i++)
		{
			if(map[u][i])
			{
				vis[i]--;
				if(vis[i]==0)
				q.push(i);
			}
		}
	}
	if(g!=0)
	printf("-1
");
	else
	{
		for(i=1;i<n;i++)
		printf("%d ",a[i]);
		printf("%d
",a[n]);
	}
}
int main()
{
	int t,o=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		getmap();		
		tuopu();
	}
	return 0;
}