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
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 .
All the weights are distrinct.
A set with m nodes 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 ,(that is, for i from 1 to m-1).For any node in the path from ),should satisfy .
Your task is to find the maximum size of Bobo Set in a given tree.
A set with m nodes 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 ,(that is, for i from 1 to m-1).For any node in the path from ),should satisfy .
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 (). Then following a line contains n integers ,all the is distrinct).Each of the following n-1 lines contain 2 integers ,denoting an edge between vertices ).
The sum of n is not bigger than 800000.
The first line contains a integer n (). Then following a line contains n integers ,all the is distrinct).Each of the following n-1 lines contain 2 integers ,denoting an edge between vertices ).
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;
}