Starship Troopers

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

 #include"iostream"
 #include"vector"
 #include"cstring"
 using namespace std;
 const int size=;
 int r,t;
 int cost[size],brain[size];
 int dp[size][size];
 vector<int> adj[size];
 void dfs(int p,int pre);
 int main()
 {
     while(cin>>r>>t)
     {
         if(r==-&&t==-)
            break;
         int bug,a,b,i;
         for(i=;i<r;i++)
         {
             cin>>bug>>brain[i];
             cost[i]=(bug+)/;
         }
         for(i=;i<r;i++)
         {
             adj[i].clear();
         }
         for(i=;i<r-;i++)
         {
             cin>>a>>b;
             adj[a-].push_back(b-);
             adj[b-].push_back(a-);
         }
         if(t==)
         {
             cout<<''<<endl;
             continue;
         }
         memset(dp,,sizeof(dp));
         dfs(,-);
         cout<<dp[][t]<<endl;
     }
     return ;
 }
 void dfs(int p,int pre)
 {
     for(int i=cost[p];i<=t;i++)
     {
         dp[p][i]=brain[p];
     }
     int num=adj[p].size();
     for(int i=;i<num;i++)
     {
         int v=adj[p][i];
         if(v==pre)
           continue;
         dfs(v,p);
         for(int j=t;j>=cost[p];j--)
           for(int k=;k<=j-cost[p];k++)
           {
                 if(dp[p][j]<dp[p][j-k]+dp[v][k])
                 {
                      dp[p][j]=dp[p][j-k]+dp[v][k];
                 }
           }
     }
 }