POJ 3687 Labeling Balls （反向拓扑排序）

最新推荐文章于 2020-03-02 00:46:01 发布

Paul-Yang

最新推荐文章于 2020-03-02 00:46:01 发布

阅读量420

点赞数

CC 4.0 BY-SA版权

分类专栏：拓扑排序 POJ-北大文章标签： POJ 3687 Labeling Balls 反向拓扑排序

本文链接：https://blog.youkuaiyun.com/shouwang_tomorrow/article/details/47731833

POJ-北大同时被 2 个专栏收录

16 篇文章

订阅专栏

拓扑排序

8 篇文章

订阅专栏

本文探讨了如何使用反向拓扑排序解决球标问题，通过限制球之间的重量关系来找到最优解决方案。包括输入解析、拓扑排序实现及输出策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12291		Accepted: 3521

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 bindicating 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

题意：

使得小的结点尽可能的排在前面：需要反向拓扑排序
如：5->6->1; 4->3->2
应让结点1尽量排在前面，再使2尽量排在前面... ...：5 6 1 4 3 2
若直接按结点从小到大拓扑排序，无法得到正解： 4 3 2 5 6 1
反向拓扑排序，按结点从大到小的顺序这样能使结点小的尽可能的排在后面：2 3 4 1 6 5，那么反过来的5 6 1 4 3 2就是所求的拓扑序列，

已AC代码：

#include<cstdio>
#include<cstring>
int n,m;
int map[300][300];
int inde[41000];
int que[300];
void topo()
{
	int i,k,top,t=0;
	for(k=n;k>=1;--k)   //k 表示重量 从高到低依次选择可以选择的最大编号
	{
		top=-1;
		for(i=n;i>=1;--i) //j 表示每次可能选出的（即出度为0）最大编号，那么该编号一定能够排在其他所有编号的后边，
		{                   //找出度为0的最大编号使其分配最大的重量，那么最后找出的解一定是最优解。
			if(inde[i]==0)  
			{
				top=i;
				break;
			}
		}
		if(top==-1)
		{
			printf("-1\n");
			return ;
		}
//		que[t++]=top;
		que[top]=k;
		inde[top]=-1;
		for(i=1;i<=n;++i)
		{
			if(map[top][i]==1)
			{
				inde[i]--;
			}
		}
	}
	printf("%d",que[1]);
	for(i=2;i<=n;++i)
		printf(" %d",que[i]);
	printf("\n");
}

int main()
{
	int T,i,a,b;
	scanf("%d",&T);
	while(T--)
	{
		memset(map,0,sizeof(map));
		memset(inde,0,sizeof(inde));
		scanf("%d%d",&n,&m);
		for(i=0;i<m;++i)
		{
			scanf("%d%d",&a,&b);
			if(map[b][a]==0)
			{
				map[b][a]=1;
				inde[a]++;
			}
		}
		topo();
	}
	return 0;
}