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  • Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

    Crazy Bobo

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 218    Accepted Submission(s): 60


    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
     


    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5325

    解题思路:反正我是智商剩余金额不足。。。


    AC代码:顺着题解思路DFS了一下= =

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <string>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <stack>
    #include <limits.h>
    using namespace std;
    typedef long long LL;
    #define y1 y234
    #define MAXN 500010 // 1e6
    int n;
    int a[MAXN];
    vector<int> edge[MAXN];
    int ans[MAXN];
    void DFS(int u) {
        ans[u] = 1;
        int len = edge[u].size();
        for(int i = 0; i < len; i++) {
            int v = edge[u][i];
            if(!ans[v]) DFS(v);
            ans[u] += ans[v];
        }
    }
    int main() {
        while(~scanf("%d", &n)) {
            memset(ans, 0, sizeof ans);
            for(int i = 1; i <= n; i++) {
                scanf("%d", &a[i]);
                edge[i].clear();
            }
            int u, v;
            for(int i = 1; i < n; i++) {
                scanf("%d%d", &u, &v);
                if(a[u] < a[v]) edge[u].push_back(v);
                else if(a[v] < a[u]) edge[v].push_back(u);
            }
            for(int i = 1; i <= n; i++) {
                if(ans[i]) continue;
                DFS(i);
            }
            int maxn = -1;
            for(int i = 1; i <= n; i++) {
                maxn = max(ans[i], maxn);
            }
            printf("%d
    ", maxn);
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/llguanli/p/8274458.html
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