hdu 1011 Starship Troopers

Starship Troopers
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8461 Accepted Submission(s): 2360


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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JGShining

//46MS    296K    1206 B    C++
//开始理解错题意了..eng还是硬伤
/*

树形DP：
    
    题意：
        典型的树形DP，派m个士兵从头结点1下去，求能获得的
    最大价值，每个节点要花费num个士兵，价值为p 
    
    思路：
     
    dp[i][j] 表示 以i为节点派j个士兵行动的最大获利
    
    DP状态转移：(u为结点，v为子节点) 
    for i:= 0~n (n为子节点个数) 
        
        for j:= m~w (m为总士兵数，w为头结点需要士兵数)
            
            for k:= 1~j-w
                
                dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]);
    
    万恶的题意...蛋蛋的忧伤.... 
      
*/ 
#include<iostream>
#include<vector>
#define N 105
using namespace std;
struct node{
    int num;
    int p;
}np[N]; //记录每个点花费及价值 
int dp[N][N]; 
vector<int>V[N];
int n,m;
void dfs(int u,int fa)
{   
    int w=np[u].num;
    for(int i=w;i<=m;i++) 
        dp[u][i]=np[u].p;
    int n0=V[u].size();
    for(int i=0;i<n0;i++){
        int v=V[u][i];
        if(v==fa) continue;
        dfs(v,u);
        for(int j=m;j>=w;j--){
            for(int k=1;k<=j-w;k++)
                dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]);
        }
    }
}
int main(void)
{
    int a,b;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++) V[i].clear();
        if(n==-1 && m==-1) break;
        for(int i=1;i<=n;i++){ 
            scanf("%d%d",&np[i].num,&np[i].p);
            np[i].num=(np[i].num+19)/20; // 1个打20个 
        }
        for(int i=1;i<n;i++){
            scanf("%d%d",&a,&b);
            V[a].push_back(b);
            V[b].push_back(a);
        }
        if(m==0){  //特殊处理 
            puts("0");continue;
        }
        dfs(1,0);
        printf("%d\n",dp[1][m]);
    }
    return 0;
}