Crazy Bobo
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 252 Accepted Submission(s): 74
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.
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.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <fstream>
#include <string>
#include <time.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) pt(x / 10);
putchar(x % 10 + '0');
}
typedef long long ll;
typedef pair<int, int> pii;
const int N = 500000 + 10;
int n;
vector<int>G[N];
pii a[N];
int w[N], r[N];
int main() {
while (cin>>n) {
for (int i = 1; i <= n; i++) {
G[i].clear();
rd(a[i].first); a[i].second = i;
w[i] = a[i].first;
}
for (int i = 1, u, v; i < n; i++) {
rd(u); rd(v); G[u].push_back(v); G[v].push_back(u);
}
sort(a + 1, a + 1 + n);
int ans = 1;
for (int i = n; i; i--) {
int id = a[i].second;
int tmp = 1;
for (auto v : G[id]) {
if (w[id] > w[v])continue;
tmp += r[v];
}
r[id] = tmp;
ans = max(ans, tmp);
}
pt(ans); puts("");
}
return 0;
}