POJ 1947 Rebuilding Roads 树形dp 难度:2

Rebuilding Roads

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9105		Accepted: 4122

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.] 

 

题意:最小多少切割次数切割出一棵P节点的子树

思路:不要以为是切掉P个点...是切出

1 dp[i][j]节点i切成j的子树所需要的最小切数

2 有两种转移,第一种切断子树,需要+1,第二种合并子树

具体看代码,注意不要互相更新

错误5次:1 胡乱提交 2 互相更新 3 忘了非根子树要切

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn=152;

const int inf=0x7ffff;

int dp[maxn][maxn];

int des[maxn];//中间缓存防止自身更新

int e[maxn][maxn];

int len[maxn];//建图

int lef[maxn];//子节点+自身个数

int n,p;

void dfs(int s){

    lef[s]=1;//自身肯定算一个,子节点还没加上

    dp[s][1]=0;//这个时候只有不切一种可能

    if(len[s]==0){return ;}//没必要刻意



    for(int i=0;i<len[s];i++){

       int t=e[s][i];

       dfs(t);

       fill(des,des+n+1,inf);//初始化缓存

       for(int k=1;k<=lef[s];k++){

            des[k]=dp[s][k]+1;//切

       }

        for(int k=1;k<=lef[s];k++){

            for(int j=1;j<=lef[t];j++){

                des[k+j]=min(dp[s][k]+dp[t][j],des[k+j]);//不切

            }

        }

       lef[s]+=lef[t];//加上这一枝

       for(int k=1;k<=lef[s];k++){

            dp[s][k]=des[k];//从缓存中取状态

       }

        dp[s][lef[s]]=0;//不需要

    }

}

int main(){

    scanf("%d%d",&n,&p);

        memset(len,0,sizeof(len));

        for(int i=1;i<=n;i++)fill(dp[i]+1,dp[i]+n+1,inf);

        for(int i=2;i<=n;i++){

            int f,t;

            scanf("%d%d",&f,&t);

            e[f][len[f]++]=t;

        }

        dfs(1);

        int ans=dp[1][p];//1是根节点分离它不需要切

        for(int i=2;i<=n;i++)ans=min(ans,dp[i][p]+1);//非根子树都要切

       // printdp();

        printf("%d\n",ans);



    return 0;

}