poj1947 Rebuilding Roads 树形dp

Rebuilding Roads

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9103		Accepted: 4120

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

题意：给你一棵树，求最少减掉几条边，使剩下的节点个数恰好为k，经典树形dp问题。

思路：用dp[I][j]表示到第i个点时共有j个点需要减掉的最小边数

状态方程：dp[k][j]=min(dp[k][j],dp[k][j-p]+dp[v][p]-1);//v为k的子节点，由于初始化时dp[k][1]已经为减掉所有子节点的情况，所以有连通时应当-1把断的边减回来。

#include <iostream>
#include <cstring>
#include <cstdio>
#define inf 0x3f3f3f3f
using namespace std;
const int N=151;
struct node
{
    int v,next;
}node[N];
int p=1,n,m,a,b,ans;
int dp[N][N],num[N],vc[N],sum[N];
void init()
{
    memset(num,0,sizeof(num));
    memset(vc,-1,sizeof(vc));
    memset(dp,inf,sizeof(dp));
    memset(sum,0,sizeof(sum));
}
void addedge(int i,int j)
{
    node[p].v=j;
    node[p].next=vc[i];
    vc[i]=p++;
    num[i]++;
}
void dfs(int k)
{
    int s=vc[k];
    sum[k]=1;
    if(s==-1){
        dp[k][1]=0;
        return;
    }
    for(;s!=-1;s=node[s].next){
        int v=node[s].v;
        dfs(v);
        sum[k]+=sum[v];
        for(int j=sum[k];j>0;j--)
            for(int p=1;p<j;p++)
                dp[k][j]=min(dp[k][j],dp[k][j-p]+dp[v][p]-1);
    }
}
int main()
{
    init();
    scanf("%d%d",&n,&m);
    for(int i=1;i<n;i++){
        scanf("%d%d",&a,&b);
        addedge(a,b);
    }
    for(int i=1;i<=n;i++)dp[i][1]=num[i];
    dfs(1);
    ans=dp[1][m];
    for(int i=2;i<=n;i++)
        ans=min(ans,dp[i][m]+1);
    printf("%d\n",ans);
    return 0;
}