POJ 3687 Labeling Balls

最新推荐文章于 2021-05-30 18:03:57 发布

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最新推荐文章于 2021-05-30 18:03:57 发布

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分类专栏：拓扑排序

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本文介绍了一种特殊的球排序问题，需要根据一系列约束条件为不同重量的球分配标签，并提供了一个具体的算法实现。该算法通过反向建图的方式进行拓扑排序，确保了在满足所有约束条件下，尽可能将较轻的球分配较小的标签。

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13926		Accepted: 4025

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

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778

因为要求小的尽量往前排，所以普通的拓扑排序就不可以了，要进行反向建图。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
int map[210][210];
int indegree[1100];
int qu[1100];
void tuopu(int n)
{
	int i,k,j;
	for(i=n;i>=1;i--)
	{
		for(j=n;j>=1;j--)
		{
			if(indegree[j]==0)
			{
				indegree[j]=-1;
				qu[j]=i;
				for(k=1;k<=n;k++)
					if(map[j][k])
						indegree[k]--;
				break;
			}	
		}
		if(j<1)
		break;
	}
    if(i>=1)
    printf("-1\n");
    else
    {
    	for(i=1;i<n;i++)
    	printf("%d ",qu[i]);
    	printf("%d\n",qu[i]);
	 } 
}
int main()
{
	int n,m,t,a,b,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		memset(map,0,sizeof(map));
		memset(indegree,0,sizeof(indegree));
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&a,&b);
			if(map[b][a]==0)
			{
				map[b][a]=1;
			    indegree[a]++;
			}
		    
		}
	    tuopu(n);
	}
	return 0;
}