PKU 1947 Rebuilding Roads

 

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



对我自己，主要要学的如果建立术的孩子兄弟表示。


动规的部分见注释吧。




//#include "stdafx.h"
//============================================================================
// Name        : acmcpp.cpp
// Author      : Yubo Chow
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <algorithm>
#include <set>
#include <limits.h>
#include <float.h>
#include <bitset>
#include <cmath>
#include <vector>
#include <map>
#include <list>
using namespace std;

int n, p;
int dp[151][151];

typedef struct
{
	int son, brother;
} node;

node nodes[151];

void solve(int s)
{
	for (int i = 0; i <= p; ++i)
	{
		dp[s][i] = 0xffffff;
	}

	dp[s][1] = 0;
	int k = nodes[s].son;

	while (k)//当拥有子树k以后，之前的子树已经算进去了
	{
		solve(k);

		for (int i = p; i >= 0; --i)//倒序可以避免k子树被重复计算
		{
			++dp[s][i];//想要得到之前的结果，需要砍掉子树k,因为之前没有算进k

			for (int j = 0; j <= i; ++j)
			{
				dp[s][i] = min(dp[s][i], dp[s][j] + dp[k][i - j]);//主要的动规方程，也就是没有k的树有j个连通，k上有i－j个，两者相连就是i个点
			}
		}

		k = nodes[k].brother;
	}
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("case.txt", "r", stdin);
#endif
	cin >> n >> p;

	bitset<151> bs;

	memset(nodes, 0, sizeof nodes);

	for (int i = 1; i < n; ++i)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		bs.set(b);
		//建立二叉树
		nodes[b].brother = nodes[a].son;
		nodes[a].son = b;
	}

	int ret = INT_MAX;

	for (int i = 1; i <= n; ++i)
	{
		if (!bs.test(i))//根节点没有入度
		{
			solve(i);
			ret = dp[i][p];
			break;
		}
	}

	for (int i = 1; i <= n; ++i)
	{
		ret = min(ret, dp[i][p] + 1);//如果不是跟节点，将其和跟节点剥离，需要增加一条断边
	}

	cout << ret << endl;
	return 0;
}