2015 多校联赛 ——HDU5325（DFS）

 

Crazy Bobo

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1215    Accepted Submission(s): 366

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 (1≤n≤500000). Then following a line contains n integers w1,w2,...,wn (1≤wi≤109,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 (1≤ai,bi≤n).
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

2015 Multi-University Training Contest 3

 

 

题意：给一个n，然后给n个点的值，再输入n-1条边，构成一个树（当时没看懂要求啥 QAQ）后来看别人的解题报告大致明白，

3    30    350    100    200    300    400    可以建成以下有向图

1 -->2-->3<--4-->5-->6-->7     当在点4的时候，能总共走过5个点，所以输出5（感觉不难  - -！！  论英语的重要性）

用深搜要手动扩栈，C++提交，否则会出现  Runtime Error  (ACCESS_VIOLATION)   //表示新手并不知道这是啥

当时就是因为提交出了这个，以为这题自己想得太简单，就放弃了 - -

而且官方给的测试数据，正确代码也只能输出一半，搞得一直以为自己错了！！

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")//手动扩栈
typedef long long ll;
using namespace std;
const int maxn= 5e5 + 5;
const ll INF = 1000000000000000000;
int p[maxn];
vector<int>q[maxn];
int num[maxn];


void dfs(int u)
{
    num[u]++;
    int len = q[u].size();
    for(int i = 0;i < len;i++)
    {
        int son = q[u][i];
        if(!num[son])
            dfs(son);
        num[u]+=num[son];
    }
}

int main()
{
    int n;
    //freopen("10.txt","r",stdin);
    while(scanf("%d",&n) != EOF)
    {
        for(int i = 1;i <= n;i++)
            scanf("%d",&p[i]);

        int a,b;
        for(int i = 1;i <= n;i++)
            q[i].clear();
        for(int i = 1;i < n;i++)
        {
            scanf("%d%d",&a,&b);
            if(p[a] < p[b])
                q[a].push_back(b);
            else
                q[b].push_back(a);
        }
        int ans = 0;
        memset(num,0,sizeof(num));
        for(int i = 1;i <= n;i++)
        {
            if(!num[i])
                dfs(i);
            ans= max(ans,num[i]);
        }

        printf("%d\n",ans);
    }
}