• HDU5325——DP+搜索——Crazy Bobo


    Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each node has a weight wi. All the weights are distrinct.
    A set with m nodes v1,v2,...,vm is a Bobo Set if:
    - The subgraph of his tree induced by this set is connected.
    - After we sort these nodes in set by their weights in ascending order,we get u1,u2,...,um,(that is,wui<wui+1 for i from 1 to m-1).For any node x in the path from ui to ui+1(excluding ui and ui+1),should satisfy wx<wui.
    Your task is to find the maximum size of Bobo Set in a given tree.

     


    Input
    The input consists of several tests. For each tests:
    The first line contains a integer n (1n500000). Then following a line contains n integers w1,w2,...,wn (1wi109,all the wi is distrinct).Each of the following n-1 lines contain 2 integers ai and bi,denoting an edge between vertices ai and bi (1ai,bin).
    The sum of n is not bigger than 800000.
     


    Output
    For each test output one line contains a integer,denoting the maximum size of Bobo Set.
     


    Sample Input
    7 3 30 350 100 200 300 400 1 2 2 3 3 4 4 5 5 6 6 7
     


    Sample Output
    5
     


    Source
     


    Recommend
    wange2014   |   We have carefully selected several similar problems for you:  5324 5323 5322 5321 5320 
    /*
    大意:找到最长的上升序列(要求连在一起)
    DP思想 dp[u] += dp[v]
    从最长的开始找上升
    */
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    using namespace std;
    int n;
    int b[500010];
    vector<int> G[500010];
    int dp[500010];
    struct edge{
        int num, id;
    }a[500010];
    
    bool cmp(edge i, edge j)
    {
        return i.num < j.num;
    }
    int main()
    {
        int x, y;
       while(~scanf("%d", &n)){
           for(int i = 1; i < n ; i++)
               G[i].clear();
        for(int i = 1; i <= n ; i++){
            scanf("%d", &a[i].num);
            a[i].id = i;
        }
        for(int i = 1; i <= n; i++)
            b[i] = a[i].num;
        sort(a + 1, a + n + 1,cmp);
        for(int i = 1; i < n ; i++){
            scanf("%d%d", &x, &y);
            G[y].push_back(x);
            G[x].push_back(y);
        }
        int max1 = 1;
        memset(dp, 0, sizeof(dp));
        for(int i = n ; i >= 1; i--){
            int u = a[i].id;
            dp[u] = 1;
            for(int j = 0 ; j < G[u].size(); j++){
                int  v = G[u][j];
                if(b[v] > b[u]) {
                    dp[u] += dp[v];
                   // printf("%d
    ", dp[u]);
            }
            }
            max1 = max(dp[u], max1);
        }
        printf("%d
    ", max1);
       }
       return 0;
    }
    

      

  • 相关阅读:
    GTC China 2016观感
    关于OpenGL的绘制上下文
    Voreen(三) 光线投射参数介绍
    分享一些DICOM数据下载网站
    Voreen (二) 入点出点计算
    Voreen (一) GPU Raycast主流程
    GPU渲染和GDI
    程序媛壮志雄心尝试装机,命运多舛壮志未酬失败告终~
    安装Newton版Glance
    安装Newton版Swift,配合keystone认证
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4685750.html
Copyright © 2020-2023  润新知