POJ-2486 Apple Tree (树上背包经典题)

最新推荐文章于 2020-08-25 11:41:28 发布

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#动态规划 #ACM #树上背包 #树形dp

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这篇博客主要探讨了ACM竞赛中的一道经典题目——POJ-2486 Apple Tree，该题目涉及到树上背包的动态规划策略。通过实例输入和输出，作者详细解释了解题思路和解决方案，旨在帮助读者掌握树形DP的应用技巧。

Apple Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11188		Accepted: 3768

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

Sample Output

11
2

#include <stdio.h>
#include <vector>
#include <string.h>
using namespace std;
int dp[101][201][2], a[101];
vector<vector<int> > g(101);
void dfs(int x, int fa, int k){
	for(int i = 0; i <= k; ++i){
		dp[x][i][0] = dp[x][i][1] = a[x];
	}
	int cur;
	for(int i = 0; i < g[x].size(); ++i){
		cur = g[x][i];
		if(cur == fa) continue;
		dfs(cur, x, k);
		for(int j = k; j >= 0; --j){
			for(int s = 1; s <= j; ++s){
				if(s >= 2){
					dp[x][j][0] = max(dp[x][j][0], dp[x][j - s][0] + dp[cur][s - 2][0]);
				}
				dp[x][j][1] = max(dp[x][j][1], dp[x][j - s][0] + dp[cur][s - 1][1]);
				if(s >= 2){
					dp[x][j][1] = max(dp[x][j][1], dp[x][j - s][1] + dp[cur][s - 2][0]);
				}
			}
		}
	}
}
int main(){
	int n, k;
	while(scanf("%d %d", &n, &k) != EOF){
		for(int i = 1; i <= n; ++i){
			g[i].clear();
		}
		for(int i = 1; i <= n; ++i){
			scanf("%d", &a[i]);
		}
		int u, v;
		for(int i = 1; i < n; ++i){
			scanf("%d %d", &u, &v);
			g[u].push_back(v);
			g[v].push_back(u);
		}
		memset(dp, 0, sizeof(dp));
		dfs(1, 0, k);
		printf("%d\n", max(dp[1][k][0], dp[1][k][1]));
	}
	return 0;
}

/*
题意：
一棵树，100个节点，每个节点上有一个或零个苹果，从1号节点出发，一共走K步，问最多可以吃多少个苹果。

思路：
树上背包的裸题。dp[i][j][0]表示在i节点，有j步可以走，最后返回i节点的情况下可以吃多少苹果，
dp[i][j][1]表示在i节点，有j步可以走，最后不返回i节点的情况下可以吃多少苹果。
先dfs到叶子节点，显然0步就可以吃到苹果，然后回溯，对父亲节点来说，所有儿子按顺序跑一遍背包，背包容量
从0到K，然后对于每个儿子节点要么返回要么不返回，区别处理一下即可。
*/