POJ 2486 Apple Tree（樹形dp）

畫船聽雨發表於2014-06-21
APP

題目的大體意思是：一個人從根節點1出發開始吃蘋果，每個節點有num[i]個蘋果。最多可以走k步，求k步以內可以吃到的蘋果的最大值。

分析：對於每一個節點都有兩種可能性存在就是：可以返回，或者不返回之前的節點。

所以有：

dp[0][i][j]  表示：不返回時，到第i個節點走了j步時得到的最大值。

dp[1][i][j]  表示：返回時，到第i個節點走了j步時得到的最大值。

這是一個樹上的分組揹包,注意：不返回時，因為只有一個“出口”走下去，所以有兩種情況：（1）這個點是“出口”；

（2）不是“出口”，所以有：

dp[1][i][j+2] = max( dp[1][i][j+2], dp[1][i][t]+dp[1][x][j-t]);

dp[0][i][j+2] = max(dp[0][i][j+2], dp[0][i][t]+dp[1][x][j-t]);//不是出口

dp[0][i][j+1] = max(dp[0][i][j+1], dp[1][i][t]+dp[0][x][j-t]);//出口


Apple Tree
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 7010   Accepted: 2335

Description

Wshxzt is a lovely girl. She likes apple very much. One day HX takes her to an apple tree. There are N nodes in the tree. Each node has an amount of apples. Wshxzt starts her happy trip at one node. She can eat up all the apples in the nodes she reaches. HX is a kind guy. He knows that eating too many can make the lovely girl become fat. So he doesn’t allow Wshxzt to go more than K steps in the tree. It costs one step when she goes from one node to another adjacent node. Wshxzt likes apple very much. So she wants to eat as many as she can. Can you tell how many apples she can eat in at most K steps.

Input

There are several test cases in the input 
Each test case contains three parts. 
The first part is two numbers N K, whose meanings we have talked about just now. We denote the nodes by 1 2 ... N. Since it is a tree, each node can reach any other in only one route. (1<=N<=100, 0<=K<=200) 
The second part contains N integers (All integers are nonnegative and not bigger than 1000). The ith number is the amount of apples in Node i. 
The third part contains N-1 line. There are two numbers A,B in each line, meaning that Node A and Node B are adjacent. 
Input will be ended by the end of file. 

Note: Wshxzt starts at Node 1.

Output

For each test case, output the maximal numbers of apples Wshxzt can eat at a line.

Sample Input

2 1 
0 11
1 2
3 2
0 1 2
1 2
1 3

Sample Output

11
2

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x7ffffff
#define PI 3.1415926535898


const int maxn = 210;

using namespace std;

int n, k;
int num[maxn];
int vis[maxn];
int dp[2][maxn][maxn];
vector<int>g[maxn], f[maxn];

void Del(int x)
{
    vis[x] = 1;
    for(int i = 0; i < g[x].size(); i++)
    {
        int y = g[x][i];
        if(vis[y])
            continue;
        f[x].push_back(y);
        Del(y);
    }
}

void dfs(int x)
{
    for(int i = 0; i <= k; i++)
        dp[0][x][i] = dp[1][x][i] = num[x];
    for(int i = 0; i < f[x].size(); i++)
    {
        int y = f[x][i];
        dfs(y);
        for(int j = k; j >= 0; j--)
        {
            for(int t = 0; t <= j; t++)
            {
                dp[0][x][j+1] = max(dp[0][x][j+1], dp[1][x][t]+dp[0][y][j-t]);//不返回
                dp[0][x][j+2] = max(dp[0][x][j+2], dp[0][x][t]+dp[1][y][j-t]);//不返回
                dp[1][x][j+2] = max(dp[1][x][j+2], dp[1][x][t]+dp[1][y][j-t]);//返回
            }
        }
    }
}

int main()
{
    while(cin >>n>>k)
    {
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i <= n; i++)
        {
            f[i].clear();
            g[i].clear();
        }
        for(int i = 1; i <= n; i++)
            cin >>num[i];
        int x, y;
        for(int i = 0; i < n-1; i++)
        {
            cin >>x>>y;
            g[x].push_back(y);
            g[y].push_back(x);
        }
        memset(vis, 0, sizeof(vis));
        Del(1);
        dfs(1);
        cout<<dp[0][1][k]<<endl;
    }
}


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