POJ 3687 Labeling Balls（拓扑排序，逆序）

最新推荐文章于 2019-02-23 14:07:21 发布

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本文介绍了一种基于逆序拓扑排序的球排序算法，用于解决给定约束条件下为不同重量的球分配唯一标签的问题。该算法确保了每个标签的球重量最小，并通过实例展示了算法的具体实现。

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Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12256		Accepted: 3510

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:

No two balls share the same label.
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

Sample Output

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




和普通拓扑排序一样，只是逆序赋值

ac代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define INF 0x7fffffff
#define MAXN 600
#define max(a,b) a>b?a:b
#define min(a,b) a>b?b:a
using namespace std;
int pri[MAXN][MAXN];
int v[MAXN];
int num[MAXN];
int n,m;
int main()
{
	int i,j,a,b,k,t,bz;
	scanf("%d",&t);
	while(t--)
	{
		bz=0;
		scanf("%d%d",&n,&m);
		memset(pri,0,sizeof(pri));
		memset(v,0,sizeof(v));
		memset(num,0,sizeof(num));
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&a,&b);
			if(pri[b][a]==0)
			{
			    pri[b][a]=1;
			    v[a]++;
			}
		}
		for(i=n;i>=1;i--)//逆序
		{
			k=-1;
			for(j=n;j>=1;j--)
			{
				if(v[j]==0)
				{
					k=j;
					num[j]=i;
					v[j]=-1;
					break;
				}
			}
			if(k!=-1)
			{
				for(j=1;j<=n;j++)
				{
					if(pri[k][j])
					{
						v[j]--;
						pri[k][j]=0;
					}
				}
		    }
		    else
		    {
		    	bz=1;
		        break;
		    }
		}
		if(bz)
		printf("-1\n");
		else
		{
			for(i=1;i<=n-1;i++)
			{
				printf("%d ",num[i]);
			}
			printf("%d\n",num[n]);
	    }
	}
	return 0;
}