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  • HDU 5325 Crazy Bobo(思路+dfs 记忆化)

    Crazy Bobo

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 612    Accepted Submission(s): 189


    Problem Description
    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
     

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    /*
    
    參考此人博客 :http://www.mamicode.com/info-detail-948802.html
    记得用c++交
    
    */
    
    
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    #include<string>
    #include<iostream>
    #include<queue>
    #include<cmath>
    #include<map>
    using namespace std;
    
    #define N 800005
    
    vector<int>g[N];
    int n;
    int ans[N];
    int a[N];
    
    int dfs(int u)
    {
        if(ans[u]) return ans[u];
        ans[u]=1;
        for(int i=0;i<g[u].size();i++)
        {
            int to=g[u][i];
            ans[u]+=dfs(to);
        }
        return ans[u];
    }
    
    int main()
    {
        int i,j;
        while(~scanf("%d",&n))
        {
            for(i=1;i<=n;i++)
                scanf("%d",&a[i]);
            for(i=1;i<=n;i++)
                g[i].clear();
            memset(ans,0,sizeof(ans));
            int u,v;
            i=n-1;
            while(i--)
            {
                scanf("%d%d",&u,&v);
                if(a[u]<a[v]) g[u].push_back(v);
                else g[v].push_back(u);
            }
            int temp=0;
            for(i=1;i<=n;i++)
            {
                temp=max(temp,dfs(i));
            }
            printf("%d
    ",temp);
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/wzzkaifa/p/6940097.html
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