poj 1947 Rebuilding Roads(树形DP)

Rebuilding Roads

Time Limit:1000MS		Memory Limit:30000K
Total Submissions:8227		Accepted:3672

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，求最少剪掉几条边使能够得到一个p个节点的树。

思路：

用dp[i][j]代表以i为根的子树要变成j个节点需要剪掉的边数。

考虑子节点的时候不用考虑它与父节点的那条边，就当不存在，那么最后求的的最小边数加1就行了。
考虑一个节点时，有两种选择，要么剪掉跟子节点相连的边，则dp[i][j] = dp[i][j]+1;

要么不剪掉，则dp[i][j] = max(dp[i][j], dp[i][k]+dp[son][j-k]);

然后随便以一条边dfs求解就行了。

为什么可以这样求解呢。

对于树上的一个节点。要么在待求解子树中。要么不在。在的话即dp[i][p]中。不然的话就肯定在其它子树上了。

详细见代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn=250;
const int INF=0x3f3f3f3f;
struct node
{
    int v;
    node *next;
} edge[maxn<<1],*head[maxn];
int n,m,cnt,dp[maxn][maxn];
void adde(int u,int v)
{
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=&edge[cnt++];
}
void dfs(int fa,int u)
{
    int i,j,v,mm=INF;
    node *p;

    dp[u][1]=0;
    for(p=head[u];p!=NULL;p=p->next)
    {
        v=p->v;
        if(v==fa)
            continue;
        dfs(u,v);
        for(i=m;i>=1;i--)//01背包。从大到小装
        {
            mm=dp[u][i]+1;//不要该子树.
            for(j=1;j<i;j++)//要该子树
                mm=min(mm,dp[u][j]+dp[v][i-j]);
            dp[u][i]=mm;
        }
    }
}
int main()
{
    int i,u,v,ans;

    while(~scanf("%d%d",&n,&m))
    {
        cnt=0;
        memset(head,0,sizeof head);
        memset(dp,0x3f,sizeof dp);
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            adde(u,v);
            adde(v,u);
        }
        dfs(1,1);
        ans=dp[1][m];
        for(i=1;i<=n;i++)
            ans=min(ans,dp[i][m]+1);//加1是因为根转移了
        printf("%d\n",ans);
    }
    return 0;
}