POJ - 1947 Rebuilding Roads 树形dp+分组背包

最新推荐文章于 2021-09-21 20:29:23 发布

原创最新推荐文章于 2021-09-21 20:29:23 发布 · 170 阅读

0 ·

CC 4.0 BY-SA版权

背包专栏收录该内容

10 篇文章

订阅专栏

本文探讨了在地震破坏后，如何通过优化道路重建策略来保护特定数量的农场仓库不受下一次地震影响的问题。采用了一种基于树状结构的算法，通过对节点进行分组背包操作，实现了对最少道路拆除数目的计算，以隔离指定数量的仓库。

The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn't have time to rebuild any extra roads, so now there is exactly one way to get from any given barn to any other barn. Thus, the farm transportation system can be represented as a tree.

Farmer John wants to know how much damage another earthquake could do. He wants to know the minimum number of roads whose destruction would isolate a subtree of exactly P (1 <= P <= N) barns from the rest of the barns.

Input

* Line 1: Two integers, N and P

* Lines 2..N: N-1 lines, each with two integers I and J. Node I is node J's parent in the tree of roads.

Output

A single line containing the integer that is the minimum number of roads that need to be destroyed for a subtree of P nodes to be isolated.

Sample Input

Sample Output

Hint

[A subtree with nodes (1, 2, 3, 6, 7, 8) will become isolated if roads 1-4 and 1-5 are destroyed.]

题解：dp[i][j] 表示以i这个节点和子代连接如果保存j个节点的话需要去除多少遍显然dp[i][1] 就是i子代的个数因为你只能从一个子代那里更新一个状态一次显然这是一个分组背包因为该节点必须要所以更新到2就可以了在最后取最小值时，若该点不是根节点需要在结果的基础上+1，因为该点还与父节点连接着

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
const int N=150+10;
vector<int> v[N];
int dp[N][N],in[N];
int n,m;
void dfs(int u)
{
	dp[u][1]=v[u].size();//
	for(int i=0;i<v[u].size();i++)
	{
		int to=v[u][i];
		dfs(to);
		for(int j=m;j>=2;j--)
			for(int k=1;k<j;k++)
				dp[u][j]=min(dp[u][j],dp[to][k]+dp[u][j-k]-1);// 这个地方 坑点 若和子代连接 那么这条边就不去了 
	}	
} 
int main()
{
	int x,y;
	while(~scanf("%d%d",&n,&m))
	{
		for(int i=1;i<=n;i++) v[i].clear();
		for(int i=1;i<n;i++)
		{
			scanf("%d%d",&x,&y);
			v[x].push_back(y);
		}
		memset(dp,INF,sizeof(dp));
		dfs(1);
		int ans=dp[1][m];
		for(int i=2;i<=n;i++) ans=min(ans,dp[i][m]+1);
		printf("%d\n",ans);
	}
	return 0;
}