POJ1947 Rebuilding Roads[树形背包]

Rebuilding Roads

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11495		Accepted: 5276

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

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题意：给定一棵节点数为n的树，问从这棵树最少删除几条边使得某棵子树的节点个数为p

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一开始想了个倒着选了几条边，其实正着也可以，先d[i][1]=子节点数量

d[i][j]表示以i为根的子树节点数为j最少删几条边

注意size要加上自己

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=155,INF=1e9;
int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}
int n,m,u,v,w,ind[N];
struct edge{
    int v,w,ne;
}e[N<<1];
int h[N],cnt=0;
void ins(int u,int v){
    cnt++;
    e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
}
int d[N][N],size[N];
void dfs(int u){
    int child=0;size[u]=1;//!self
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v;
        dfs(v);
        size[u]+=size[v];
        child++;
    }
    if(!child) {size[u]=1;d[u][1]=0;return;}//printf("size %d %d
",u,size[u]);
    
    d[u][1]=child;
    for(int j=2;j<=size[u];j++) d[u][j]=INF;
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v;
        for(int j=size[u];j>=1;j--){
            int t=min(j-1,size[v]);
            for(int k=1;k<=t;k++) d[u][j]=min(d[u][j],d[u][j-k]+d[v][k]-1);
        }
    }
    
    //for(int i=1;i<=size[u];i++) printf("d %d %d %d
",u,i,d[u][i]);
}
int main(int argc, const char * argv[]) {
    n=read();m=read();
    for(int i=1;i<=n-1;i++){
        u=read();v=read();ins(u,v);ind[v]++;
    }
    int root=-1;
    for(int i=1;i<=n;i++) if(!ind[i]) {root=i;break;}
    dfs(root);
    int ans=INF;
    for(int i=1;i<=n;i++) if(size[i]>=m) ans=min(ans,d[i][m]+(i==root?0:1));
        //,printf("ans %d %d
",i,d[i][m]);
    printf("%d",ans);
    return 0;
}