多校 hdu 5325

Crazy Bobo

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1308    Accepted Submission(s): 400

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

73 30 350 100 200 300 4001 22 33 44 55 66 7

 

Sample Output

5

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cmath>#include<stdlib.h>#include<map>#include<set>#include<time.h>#include<vector>#include<queue>#include<string>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define eps 1e-8#define INF 0x3f3f3f3f#define LL long long#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))#define maxn 500000 + 10int w[maxn], a[maxn];struct edge{    int v, next;}e[maxn * 2];int n;int vis[maxn];int head[maxn], tot;void init(){    tot = 0;    memset(vis, 0, sizeof(vis));    memset(a, 0, sizeof(a));    memset(head, -1, sizeof(head));}void adde(int u, int v){    e[tot].v = v;    e[tot].next = head[u];    head[u] = tot++;}void dfs(int u){    vis[u] = 1;    a[u] = 1;    for(int i=head[u]; i!=-1; i=e[i].next)    {        int v = e[i].v;        //printf("%d--%d\n", u, v);        if(!vis[v])            dfs(v);        a[u] += a[v];    }}int main(){    while(~scanf("%d", &n))    {        init();        for(int i=1; i<=n; i++)            scanf("%d", w + i);        for(int i=1; i<n; i++)        {            int u, v;            scanf("%d%d", &u, &v);            if(w[u] < w[v])                adde(u, v);            else if(w[u] > w[v])                adde(v, u);        }        for(int i=1; i<=n; i++)            if(!vis[i])            dfs(i);        int ans = -1;        for(int i=1; i<=n; i++)            ans = max(ans, a[i]);        printf("%d\n", ans);    }    return 0;}