Rebuilding Roads
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 10653 Accepted: 4884

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

題目鏈接：
http://poj.org/problem?id=1947

題意：
給你一棵節點為n的樹。問至少砍幾刀能夠孤立出一棵節點為m的子樹。

思路：
找到根節點（入度為０）後ｄｆｓ在每一個節點統計dp[i][j]
表示ｉ及其的子樹保留ｊ個節點的最小代價；

int tmp=dp[x][ii]+1;//不要ｙ節點；
tmp=min(tmp,dp[x][ii-j]+dp[y][j]);
dp[x][ii]=tmp;

ｙ為ｘ的兒子節點。

代碼：

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<vector>
using namespace std;

int dp[155][155];
int ans;
int in[155];
vector<int> lin[155];
int n,aa,bb,p;
void dfs(int x)
{
    for(int i=2;i<=p;i++) dp[x][i]=99999999;
    if(lin[x].size()==0)
    dp[x][1]=0;
    for(int i=0;i<lin[x].size();i++)
    {
        int y=lin[x][i];
        dfs(y);
        for(int ii=p;ii>=1;ii--)
        {
            int tmp=dp[x][ii]+1;
            for(int j=1;j<ii;j++)
            tmp=min(tmp,dp[x][ii-j]+dp[y][j]);
            dp[x][ii]=tmp;
        }       
    }
}
int main()
{
    scanf("%d%d",&n,&p);
    {
        int ans=99999999;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&aa,&bb);
            lin[aa].push_back(bb);
            in[bb]++;
        }
        for(int i=1;i<=n;i++)
        if(!in[i])
        {
            dfs(i);
            ans=dp[i][p];
            break;
        }

        for(int i=1;i<=n;i++)
        ans=min(ans,dp[i][p]+1);
        printf("%d\n",ans);
    }

}

[poj 1947] Rebuilding Roads 樹形ＤＰ