HDU：1011 Starship TroopersStarship Troopers

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概述

Starship Troopers

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17993 Accepted Submission(s): 4767

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

Sample Output


50
7
题目意思：有N个房间，有M个士兵，一个士兵可以消除20个bugs,每个房间中有一定数目的bugs和一定数目的brains,某些房间有隧道相连是相通的，
接下来给出N个房间的bugs数，和brains数，接下来N-1组数据给出哪两个房间是相通的。第一个房间为入口，求给你N个房间，M个士兵，以及每个房间
的bugs数和brains数，从1进能获得的最大brains数目。
解题思路：
好难的题目，树形dp,用的是dfs+dp
题目代码：
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#define max(a,b) a>b? a:b
using namespace std;
int N,M;
int dp[102][102];
int vis[102];
vector<int> room[103];//用于存储图
int bugs[103];
int brains[103];
void dfsandtreedp(int child,int father)
{
    for(int i = bugs[child]; i <= M; i++)
    {
        dp[child][i] = brains[child];
    }
    int ssize = room[child].size();//计算与child相通的房间有几个
    vis[child] = 1;//代表child访问过了
    for(int i = 0; i < ssize; i++)//对与它相邻的房间进行搜索
    {
       int roomnum;
       roomnum = room[child][i];
       if(!vis[roomnum])//如果与它相邻的房间没有访问，就访问之，以roomnum为跟进行深搜
        dfsandtreedp(roomnum,child);//roomnum变为child,child变成father
    }//走到这里，代表与child所有相通的房间都访问过了，那么child就是叶子节点，他的父节点为father.
    for(int i = M; i >= bugs[father]; i--)
    {
       for(int j = 1; j <= i - bugs[father]; j++)
            dp[father][i] = max(dp[father][i],dp[father][i-j]+dp[child][j]);
    }
    /*dp[father][i] = max(dp[father][i],dp[father][i-j]+dp[child][j]; dp[father][i]
   是以father为跟节点，士兵数为i，时所能获得的最大brains数目，father会用去bugs[father]个
   个士兵，则会给他的孩子剩下i - bugs[father]个士兵，
}
int main()
{
    while(~scanf("%d%d",&N,&M))//N为房间数量，M为士兵数量
    {
        if(N == -1 && M == -1)//结束条件
            break;
        memset(dp,0,sizeof(dp));//dp刷为0
        memset(vis,0,sizeof(vis));//vis初始为0，代表所有节点都未访问
        for(int i = 1; i <= N; i++)
        {
            scanf("%d%d",&bugs[i],&brains[i]);//输入每个房间的bugs数，和brains数
            bugs[i] = (bugs[i] + 19)/20;//这一步是转换成需要多少个兵
        }
        for(int i = 1; i <= N; i++)
        {
            room[i].clear();//清空容器
        }
        for(int i = 1; i < N; i++)//输入n-1组连通的房间
        {
            int temp1,temp2;
            scanf("%d%d",&temp1,&temp2);
            room[temp1].push_back(temp2);
            room[temp2].push_back(temp1);
        }
        if(M == 0)//提供兵的数目为0，直接输出0，结束
        {
            printf("0n");
            continue;
        }
        dfsandtreedp(1,0);
        printf("%dn",dp[1][M]);//以1为根节点，兵为M时所得的最大brains数目
    }
    return 0;
}