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

/*

题意:起点1，然后n个点，m个人，每一个节点有 cost 和va,一个人最多消灭20个cost
     但是不能走一个节点两遍，消灭一个节点的全部cost后得到这个节点的va,
     而且消灭他的人不能再用了，求这m个人最多可以得到多少 va

思路: 树形dp+背包，背包部分在代码写出

*/

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

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define INF 0x3f3f3f3f
#define N 155

vector<int>g[N];

int dp[N][N];
int cost[N],va[N];
int n,m;
int vis[N];

void dfs(int u)
{
    int i,j;
    int lim=(cost[u]+19)/20;
    for(i=lim;i<=m;i++)
	   dp[u][i]=va[u];

	vis[u]=1;
	for(i=0;i<g[u].size();i++)
	{
	     int to=g[u][i];
	     if(vis[to]) continue;
	     dfs(to);
	     for(int v=m;v>=lim;v--)     //假设这个点的人数最大为m,
	                                 // 那么依次求人数在m~lim时的最大情况(背包)
			for(int co=1;v-co>=lim;co++)   //枚举分几个人给to节点
			  dp[u][v]=max(dp[u][v],dp[u][v-co]+dp[to][co]);
	}
}

int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m))
	{
		if(n==-1&&m==-1) break;
		for(i=1;i<=n;i++)
			scanf("%d%d",&cost[i],&va[i]);
		for(i=0;i<=n;i++)
			g[i].clear();

		int u,v;
		for(i=1;i<n;i++)
		{
			 scanf("%d%d",&u,&v);
		     g[u].push_back(v);
		     g[v].push_back(u);
		}
		if(m==0)     //这个地方很重要，对于起点没有cost的情况,没有这个会挂
		{
			printf("0\n");
			continue;
		}
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        dfs(1);
	   printf("%d\n",dp[1][m]);
	}
   return 0;
}

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