Starship Troopers （樹形dp+背包）

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      

題目大概：

有一顆有根樹，每個結點都有價值和敵人數，必須派士兵擊敗敵人，才能占領這個結點，取得價值。問k個士兵能獲得的最大價值。

思路：

dp[i][j]指的是，第i個節點，派出j個士兵獲得的價值。

周遊一遍數，回來的時候順便做一下dp背包。利用子節點更新一下本結點的資訊。

狀态轉移方程    dp[x][o]=max(dp[x][o],dp[x][o-j]+dp[son][j]);   到本節點，派出的o個士兵，可以在到達本節點之前派出，也可以在本結點之後派出。

發現還是利用結構體，存儲樹好，一般也不會逾時。

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int n,p;
int dp[105][105];
int d[105],nao[105];
int head[105];
int minshu,no;
int ans;
struct shu
{
    int v;
    int next;
}tr[210];

void add(int q,int w)
{
    tr[ans].v=w;
    tr[ans].next=head[q];
    head[q]=ans++;

}

void dfs(int x,int pa)
{
    int bing=ceil((double)d[x]/20.0);
    for(int i=bing;i<=p;i++)dp[x][i]=nao[x];


    for(int i=head[x];i!=-1;i=tr[i].next)
    {
        int son=tr[i].v;
        if(son!=pa)
        {
            dfs(son,x);
            for(int o=p;o>=bing;o--)
            {
                for(int j=1;j<=o-bing;j++)
                {
                    dp[x][o]=max(dp[x][o],dp[x][o-j]+dp[son][j]);
                }
            }

        }
    }


}

int main()
{


   while(scanf("%d%d",&n,&p))
   {
       if(n==-1&&p==-1)break;
       memset(dp,0,sizeof(dp));
       memset(d,0,sizeof(d));
       memset(head,-1,sizeof(head));
       memset(nao,0,sizeof(nao));
       ans=0;
       for(int i=1;i<=n;i++)
       {
           scanf("%d%d",&d[i],&nao[i]);
       }

       for(int i=1;i<n;i++)
       {
           int q,w;
            scanf("%d%d",&q,&w);
            add(q,w);
            add(w,q);

       }
       if(p==0){printf("0\n");continue;}
       dfs(1,0);

       printf("%d\n",dp[1][p]);
   }
    return 0;
}