Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
2018-01-12 10:15
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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 (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
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\n", maxn); } return 0; }
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