poj1947 Rebuilding Roads--树形dp

原题链接：​​http://poj.org/problem?id=1947​​

题意：给出n,p，一共有n个节点，要求最少减去最少的边是多少，剩下p个节点
思路：典型的树形DP，
dp[s][i]:记录s结点，要得到一棵j个节点的子树去掉的最少边数
 考虑其儿子k
 1）如果不去掉k子树，则
 dp[s][i] = min(dp[s][j]+dp[k][i-j])  0 <= j <= i

 2）如果去掉k子树，则
 dp[s][i] =  dp[s][i]+1
 总的为
 dp[s][i] = min (min(dp[s][j]+dp[k][i-j]) ,  dp[s][i]+1 )

#define _CRT_SECURE_NO_DEPRECATE 

#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#include<string>
#include<stdio.h>
#define INF 99999999
#define eps 0.0001
using namespace std;

int n, p;
int father[155];
int dp[155][155];//dp[i][j]表示以i为树根分割成k子树的最小数目
vector<int> V[155];

void tree_dp(int fat)
{
  for (int i = 0; i < V[fat].size(); i++)
  {
    int son = V[fat][i];
    tree_dp(son);
    for (int j = p; j > 1; j--)//对于root来说
      for (int k = 1; k < j; k++)
        dp[fat][j] = min(dp[fat][j], dp[fat][j - k] + dp[son][k] - 2);//对于fat，已经把儿子son的连线加上；对于儿子son也把父亲fat连线加进去，所以是减2，不是减1
  }
}

int main() 
{
  while (~scanf("%d%d", &n, &p))
  {
    
    for (int i = 1; i <= n; i++)
    {
      father[i] = 0;
      V[i].clear();
    }

    int u, v;
    for (int i = 1; i < n; i++)
    {
      scanf("%d%d", &u, &v);
      father[v] = u;
      V[u].push_back(v);
    }

    for (int i = 1; i <= n; i++)
    {
      dp[i][1] = V[i].size() + 1;
      for (int j = 2; j <= p; j++)
        dp[i][j] = INF;
    }

    int root = 1;
    while (father[root])
      root = father[root];

    tree_dp(root);
    dp[root][p]--;
    int ans = INF;
    for (int i = 1; i <= n; i++)
      ans = min(ans, dp[i][p]);
    printf("%d\n", ans);
  }
  return 0;
}