HDU1011 树形DP

Starship Troopers

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17560    Accepted Submission(s): 4659

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：

树上有m个房间，每个房间里有bug和价值，一个士兵可以消灭20个bug，只有消灭掉房间里的所有bug才能得到价值，不能回到走过的房间现在有k个士兵问最多可以得到多少价值。

代码：

 /*
 DP方程不好理解。不足20个bug也要安排士兵。
 */
 #include<iostream>
 #include<string>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<iomanip>
 #include<queue>
 #include<stack>
 using namespace std;
 int n,m,lne;
 int dp[][];
 int head[];
 int a[],b[];
 bool vis[];
 struct node
 {
     int to,next;
 }tree[];
 void add(int u,int v)
 {
     tree[lne].to=v;
     tree[lne].next=head[u];
     head[u]=lne++;
 }
 void dfs(int root)
 {
     vis[root]=;
     for(int i=a[root];i<=m;i++)
     dp[root][i]=b[root];//能获得b的情况下，赋值
     for(int i=head[root];i!=-;i=tree[i].next)
     {
         int son=tree[i].to;
         if(vis[son])
         continue;
         dfs(son);
         for(int j=m;j>=a[root];j--)
         {
             for(int k=;k+j<=m;k++)
             dp[root][j+k]=max(dp[root][j+k],dp[root][j]+dp[son][k]);//一共带有j+k个士兵在root点放j个士兵
                                                                    //dp[root][j+k]是在士兵数量够用的情
             //况下的值，当这个父亲当儿子时他就有两个值一个是此值，一个是0，当dp[son][k]中的k大于等于他
             //需要的士兵时dp[son][k]不是0，k小于他需要的士兵时dp[son][k]取0。
         }
     }
 }
 int main()
 {
     int u,v;
     while(scanf("%d%d",&n,&m))
     {
         if(n==-&&m==-)
         break;
         memset(dp,,sizeof(dp));
         memset(head,-,sizeof(head));
         memset(vis,,sizeof(vis));
         for(int i=;i<=n;i++)
         {
             scanf("%d%d",&a[i],&b[i]);
             a[i]=(a[i]+)/;
         }
         lne=;
         for(int i=;i<n-;i++)
         {
             scanf("%d%d",&u,&v);
             add(u,v);
             add(v,u);
         }
         if(m==) //没有士兵不能得到价值
         {
             printf("0\n");
             continue;
         }
         dfs();
         printf("%d\n",dp[][m]);
     }
     return ;
 }