HDU1011 Starship Troopers(树形dp)

Starship Troopers

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15816    Accepted Submission(s): 4231

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 1050 1040 1040 2065 3070 301 21 32 42 51 120 7-1 -1

 

Sample Output

507

 题意：给你M个士兵进一幅树形地图里消灭bugs,一个有N个房间，每个房间有一定bugs和大脑存在的可能性，每间房间都必须留下士兵消灭bugs并且留下的士兵不能再走，即使bugs为0也必须在至少有一个士兵的情况下才能通过（不用留下士兵），不能回到先前已经经过的房间。问如何走使得房间里的大脑可能性加起来最大。

思路：树形dp,一开始还认为只能往一个方向走，没想到原来可以兵分多路，那么问题就可以变为每条路分配多少士兵使得最后能得到最大值。

dp[fa][j] = max(dp[fa][j], dp[fa][j-k]+dp[son][k]);dp[i][j]表示第i个点分配j个士兵能得到的最大值，最后输出dp[1][M]。

注意：1是起点但不一定是根结点，所以要建无向图。

#include <stdio.h>#include <string.h>#include <algorithm>#define maxn 105using namespace std;int dp[maxn][maxn], room[maxn][2];int head[maxn], next[maxn];bool visfa[maxn];struct node{    int to, next;}edge[maxn*2];int N, M, tot;void add(int u, int v){    edge[tot].to = v;    edge[tot].next =  head[u];    head[u] = tot++;}void dfs(int fa){    //int ans = 0;    visfa[fa] = 1;    int i, j, k, son, tmp;    for(i = head[fa];i != -1;i = edge[i].next){        son = edge[i].to;        //printf("fa:%d  son:%d\n", fa, son);        if(!visfa[son]) dfs(son);        else continue;        tmp = room[fa][0];        for(j = M;j >= tmp;j--){//之所以j--，是因为需要的是未被更新的状态。            for(k = 1;k <= j - tmp;k++)                dp[fa][j] = max(dp[fa][j], dp[fa][j-k]+dp[son][k]);        }    }}int main(){    int i, u, v, j, tmp;    while(scanf("%d %d", &N, &M) != EOF)    {        if(N==-1&&M==-1) break;        tot = 1;        memset(head, -1, sizeof head);        memset(dp, 0, sizeof dp);        memset(visfa, 0, sizeof visfa);        for(i = 1;i <= N;i++){            scanf("%d %d", &room[i][0], &room[i][1]);            tmp = (room[i][0]+19)/20;            room[i][0] = tmp;        }        for(i = 1;i <= N-1;i++){            scanf("%d %d", &u, &v);            add(u, v);            add(v, u);        }        for(i = 1;i <= N;i++)            for(j = room[i][0];j <= M;j++)                dp[i][j] = room[i][1];        dfs(1);        if(M==0) printf("0\n");        else printf("%d\n", dp[1][M]);    }}