Crazy Bobo
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1334 Accepted Submission(s): 410
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.
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.
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
Author
ZSTU
Source
解题:直接搜索。。。建立有向图时候,小权向大权的连边,然后看看每个每个点,最多能走多少个点。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <vector> 5 #include <cstring> 6 #pragma comment(linker, "/stack:1024000000,1024000000") 7 using namespace std; 8 const int maxn = 500010; 9 vector<int>g[maxn]; 10 int w[maxn],ret[maxn]; 11 void dfs(int u) { 12 ret[u] = 1; 13 for(int i = g[u].size()-1; i >= 0; --i) { 14 if(!ret[g[u][i]]) dfs(g[u][i]); 15 ret[u] += ret[g[u][i]]; 16 } 17 } 18 int main() { 19 int n,u,v; 20 while(~scanf("%d",&n)){ 21 for(int i = 1; i <= n; ++i){ 22 scanf("%d",w+i); 23 g[i].clear(); 24 } 25 for(int i = 1; i < n; ++i){ 26 scanf("%d%d",&u,&v); 27 if(w[u] < w[v]) g[u].push_back(v); 28 else g[v].push_back(u); 29 } 30 memset(ret,0,sizeof ret); 31 int ans = 0; 32 for(int i = 1; i <= n; ++i){ 33 if(!ret[i]) dfs(i); 34 ans = max(ans,ret[i]); 35 } 36 printf("%d ",ans); 37 } 38 return 0; 39 }