sample tor http ber uil tran isdigit ast ext

[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 題目大意：有一顆N個節點的樹，問最少刪去幾條邊使剩下的樹的大小有一顆為P？ 直接寫不就好了麽？ 代碼：

#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號節點的子樹中隔離成為大小為j個的道路數量
//dp[i][k]=min(dp[i->son][j]+dp[i][k-j]) 

【樹形dp】Rebuilding Roads