【树形dp】Rebuilding Roads

[POJ1947]Rebuilding Roads

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11934		Accepted: 5519

Description

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

11 6
1 2
1 3
1 4
1 5
2 6
2 7
2 8
4 9
4 10
4 11

Sample Output

2

Hint

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

Source

USACO 2002 February

 

题目大意：有一颗N个节点的树，问最少删去几条边使剩下的树的大小有一颗为P？

直接写不就好了么？

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN=100001;
const int INF=999999;
int N,M;
vector<int> vec[201];
int dp[201][201];
int ans=INF;

void dfs(int x,int fa){
	int cnt=0;
	for(int i=0;i<vec[x].size();i++){
		if(vec[x][i]!=fa)
			dfs(vec[x][i],x),cnt++;
	}
	dp[x][1]=dp[x][0]=0;
	for(int i=0;i<vec[x].size();i++){
		if(vec[x][i]!=fa)
		for(int j=M;j>=1;j--){
			if(dp[x][j]!=INF) dp[x][j]++;
			for(int k=1;k<=M;k++){
			    if(k>=j||dp[vec[x][i]][k]==INF) break;
			    if(dp[x][j-k]!=INF) dp[x][j]=min(dp[vec[x][i]][k]+dp[x][j-k],dp[x][j]);
			}
		}
	}
	if(x!=1) ans=min(ans,dp[x][M]+1);
	else ans=min(ans,dp[x][M]);
	return ;
}

int main(){
	N=read(),M=read();
    for(int i=0;i<=N;i++)
	    for(int j=0;j<=M;j++) dp[i][j]=INF;
	for(int i=1;i<N;i++){
		int u=read(),v=read();
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	dp[1][1]=0;
	dfs(1,-1);
	if(ans!=INF) printf("%d\n",ans);
	else puts("0");
}
//dp[i][j]表示i号节点的子树中隔离成为大小为j个的道路数量
//dp[i][k]=min(dp[i->son][j]+dp[i][k-j])