HDOJ-1011

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

---

思路：
首先提炼一下题意：
G = (V, E), \(f : V \rightarrow N\), \(g : V \rightarrow N\) (其中f为bugs的函数，g为possibility的函数), 给定$ m \in N^+$

设函数\(L(v_i) = (f(v_i)+19)/20\) (if \(L(v_i) == 0\) && $g(v_i) $ != 0, \(L(v_i) = 1\))

在条件\(\sum L(v_i) <= m\)下，求max{\(\sum g(v_i)\)}

看上去像是图问题（树问题）与0-1背包问题的交集。

所以思路自然而然的是动态规划。

利用0-1背包问题的思路：

设\(res[a][b]\)为以a为根的子树，在给定b下的最优解，则

\(res[a][b]\) = max(\(res[a][b]\), \(res[a][k]+res[c][b-k]\))( (a, c)\(\in E\) )

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <memory.h>
#define MAX_SIZE 110
using namespace std;
int N;
int M;
int graph[MAX_SIZE][MAX_SIZE];
int res[MAX_SIZE][MAX_SIZE];
bool visited[MAX_SIZE];
typedef struct room
{
    int bugs;
    int brains;
    int cost;
}Room;
Room rooms[MAX_SIZE];
void dp(int root)
{
    visited[root] = true;
    int r = rooms[root].cost;

    for (int i = M; i >= r; i--)
    {
        res[root][i] = rooms[root].brains;
    }

    for (int i = 1; i <= N; i++)
    {
        if (graph[root][i] && !visited[i])
        {
            dp(i);
            for (int j = M; j >= r; j--)
            {
                for (int k = 1; k <= j - r; k++)
                {
                    res[root][j] = max(res[root][j], res[root][j - k] + res[i][k]);
                }
            }
        }
    }
}
int main()
{
    while (scanf("%d %d", &N, &M) != EOF)
    {
        if (N == -1 && M == -1) break;

        memset(res, 0, sizeof(res));
        memset(visited, false, sizeof(visited));
        memset(graph, 0, sizeof(graph));

        for (int i = 1; i <= N; i++)
        {
            scanf("%d %d", &rooms[i].bugs, &rooms[i].brains);
            rooms[i].cost = (rooms[i].bugs + 19) / 20;
            if (rooms[i].cost == 0 && rooms[i].brains) rooms[i].cost = 1;
        }

        for (int i = 1; i <= N - 1; i++)
        {
            int x, y;
            scanf("%d %d", &x, &y);
            graph[x][y] = 1;
            graph[y][x] = 1;
        }

        if (M == 0)
        {
            printf("0\n");
        }
        else
        {
            dp(1);
            printf("%d\n", res[1][M]);
        }
    }
    return 0;
}

PS：我感觉我的思路是对的，但是就是AC不能， 怀疑有什么边界条件漏了。