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 }