Starship Troopers

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

You, the leader of Starship Troopers, are sent to destroy a base of the bugs. The base is built underground. It is actually a huge cavern, which consists of many rooms connected with tunnels. Each room is occupied by some bugs, and
their brains hide in some of the rooms. Scientists have just developed a new weapon and want to experiment it on some brains. Your task is to destroy the whole base, and capture as many brains as possible.

To kill all the bugs is always easier than to capture their brains. A map is drawn for you, with all the rooms marked by the amount of bugs inside, and the possibility of containing a brain. The cavern‘s structure is like a tree in such a way that there is
one unique path leading to each room from the entrance. To finish the battle as soon as possible, you do not want to wait for the troopers to clear a room before advancing to the next one, instead you have to leave some troopers at each room passed to fight
all the bugs inside. The troopers never re-enter a room where they have visited before.

A starship trooper can fight against 20 bugs. Since you do not have enough troopers, you can only take some of the rooms and let the nerve gas do the rest of the job. At the mean time, you should maximize the possibility of capturing a brain. To simplify the
problem, just maximize the sum of all the possibilities of containing brains for the taken rooms. Making such a plan is a difficult job. You need the help of a computer.

Input

The input contains several test cases. The first line of each test case contains two integers N (0 < N <= 100) and M (0 <= M <= 100), which are the number of rooms in the cavern and the number of starship troopers you have, respectively.
The following N lines give the description of the rooms. Each line contains two non-negative integers -- the amount of bugs inside and the possibility of containing a brain, respectively. The next N - 1 lines give the description of tunnels. Each tunnel is
described by two integers, which are the indices of the two rooms it connects. Rooms are numbered from 1 and room 1 is the entrance to the cavern.

The last test case is followed by two -1‘s.

Output

For each test case, print on a single line the maximum sum of all the possibilities of containing brains for the taken rooms.

Sample Input

5 10
50 10
40 10
40 20
65 30
70 30
1 2
1 3
2 4
2 5
1 1
20 7
-1 -1

Sample Output

50
7

Author

XU, Chuan

Source

ZJCPC2004

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| Note

树形dp, dp[u][i]表示派i个士兵去攻打u这个子树所能得到的最优解

dp[u][i] = max(dp[v][j] + dp[u][i - j])

/*************************************************************************
    > File Name: hdu1011.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年01月03日 星期六 14时41分19秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 150;
int dp[N][N];
int pro[N];
int bug[N];
int n, m;
vector <int> edge[N];

void dfs(int u, int fa)
{
	int tmp = bug[u] / 20;
	if (bug[u] % 20 && bug[u])
	{
		++tmp;
	}
	for (int i = tmp; i <= m; ++i)
	{
		dp[u][i] = pro[u];
	}
	int size = edge[u].size();
	for (int i = 0; i < size; ++i)
	{
		int v = edge[u][i];
		if (v == fa)
		{
			continue;
		}
		dfs(v, u);
		for (int j = m; j >= tmp; --j) //bug[u] 用来 制服u节点
		{
			for (int k = 1; k <= j - tmp; ++k) //注意k要从1开始,因为每个房间都至少有一个士兵
			{
				dp[u][j] = max(dp[u][j], dp[v][k] + dp[u][j - k]);
			}

		}
	}
}

int main()
{
	int u, v;
	while (~scanf("%d%d", &n, &m))
	{
		if (n == -1 && m == -1)
		{
			break;
		}
		memset (dp, 0, sizeof(dp));
		for (int i = 1; i <= n; ++i)
		{
			edge[i].clear();
			scanf("%d%d", &bug[i], &pro[i]);
		}
		for (int i = 1;  i <= n - 1; ++i)
		{
			scanf("%d%d", &u, &v);
			edge[u].push_back(v);
			edge[v].push_back(u);
		}
		if (m == 0)
		{
			printf("0\n");
			continue;
		}
		dfs(1, -1);
		printf("%d\n", dp[1][m]);
	}
	return 0;
}