zoukankan      html  css  js  c++  java
  • POJ 2486 Apple Tree

    好抽象的树形DP。。。。。。。。。

    Apple Tree
    Time Limit: 1000MSMemory Limit: 65536K
    Total Submissions: 6411Accepted: 2097

    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 3Sample Output

    11
    2

    Source

    POJ Contest,Author:magicpig@ZSU



    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>

    using namespace std;

    vector<int> g[200];
    int N,K,valu[200],dp[2][200][300];
    bool vis[200];

    void Tree_Dp(int s)
    {
        vis[s]=true;
        for(int i=0;i<=K;i++)
            dp[0][s]=dp[1][s]=valu[s];
        for(int i=0;i<g[s].size();i++)
        {
            int t=g[s];
            if(vis[t]) continue;
            Tree_Dp(t);
            for(int j=K;j>=0;j--)
            {
                for(int k=0;k<=j;k++)
                {
                    dp[0][s][j+2]=max(dp[0][s][j+2],dp[0][t][k]+dp[0][s][j-k]);
                    dp[1][s][j+2]=max(dp[1][s][j+2],dp[0][t][k]+dp[1][s][j-k]);
                    dp[1][s][j+1]=max(dp[1][s][j+1],dp[1][t][k]+dp[0][s][j-k]);
                }
            }
        }
    }

    int main()
    {
        while(scanf("%d%d",&N,&K)!=EOF)
        {
            for(int i=1;i<=N;i++)
            {
                scanf("%d",valu+i);
                g.clear();
            }
            for(int i=0;i<N-1;i++)
            {
                int a,b;
                scanf("%d%d",&a,&b);
                g[a].push_back(b);
                g.push_back(a);
            }
            memset(vis,false,sizeof(vis));
            memset(dp,0,sizeof(dp));
            Tree_Dp(1);
            printf("%d ",max(dp[1][1][K],dp[0][1][K]));
        }
        return 0;
    }
    * This source code was highlighted by YcdoiT. ( style: Codeblocks )

  • 相关阅读:
    Spring Data JPA简介 Spring Data JPA特点
    JavaScript的事件循环机制总结 eventLoop
    什么是 MyBatis?
    如果你也用过 struts2.简单介绍下 springMVC 和 struts2 的区别有哪些?
    SpringMvc 框架
    线程、并发、并行、进程是什么,以及如何开启新的线程?
    面试:你最大的长处和弱点分别是什么?这些长处和弱点对你在企业的业绩会有什么样的影响?
    面向对象三大特性
    一台客户端有三百个客户与三百个客户端有三百个客户对服务器施压,有什么区别?
    面向对象三大特性
  • 原文地址:https://www.cnblogs.com/CKboss/p/3350822.html
Copyright © 2011-2022 走看看