hdu 5325 Crazy Bobo 乱搞+搜索

最新推荐文章于 2018-10-14 18:03:04 发布

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最新推荐文章于 2018-10-14 18:03:04 发布

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原文链接：http://www.cnblogs.com/gaoxiang36999/p/4684200.html

本文介绍了一道算法题目，要求在给定的树结构中寻找最大规模的BoboSet，即满足特定条件的节点集合。通过DFS遍历树结构，并采用特定策略筛选节点来解决问题。

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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：反正我是智商余额不足。。。

AC代码：顺着题解思路DFS了一下= =

1 #pragma comment(linker, "/STACK:1024000000,1024000000")
2 #include <iostream>
3 #include <cstdio>
4 #include <cstring>
5 #include < string>
6 #include <cmath>
7 #include <algorithm>
8 #include <vector>
9 #include <queue>
10 #include < set>
11 #include <map>
12 #include <stack>
13 #include <limits.h>
14 using namespace std;
15 typedef long long LL;
16 #define y1 y234
17 #define MAXN 500010 // 1e6
18 int n;
19 int a[MAXN];
20 vector< int> edge[MAXN];
21 int ans[MAXN];
22 void DFS( int u) {
23     ans[u] = 1;
24      int len = edge[u].size();
25      for( int i = 0; i < len; i++) {
26          int v = edge[u][i];
27          if(!ans[v]) DFS(v);
28         ans[u] += ans[v];
29     }
30 }
31 int main() {
32      while(~scanf( " %d ", &n)) {
33         memset(ans, 0, sizeof ans);
34          for( int i = 1; i <= n; i++) {
35             scanf( " %d ", &a[i]);
36             edge[i].clear();
37         }
38          int u, v;
39          for( int i = 1; i < n; i++) {
40             scanf( " %d%d ", &u, &v);
41              if(a[u] < a[v]) edge[u].push_back(v);
42              else if(a[v] < a[u]) edge[v].push_back(u);
43         }
44          for( int i = 1; i <= n; i++) {
45              if(ans[i]) continue;
46             DFS(i);
47         }
48          int maxn = - 1;
49          for( int i = 1; i <= n; i++) {
50             maxn = max(ans[i], maxn);
51         }
52         printf( " %d\n ", maxn);
53     }
54      return 0;

55 }

转载于:https://www.cnblogs.com/gaoxiang36999/p/4684200.html