E - Preorder Test

思路:想到二分答案了之后就不难啦, 对于每个答案用树形dp取check, 如果二分的值是val, dp[ i ]表示 i 这棵子树答案不低于val的可以访问的

最多节点, 第二次dfs求出以每个点为根的答案。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std; const int N = 2e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-; int n, k, a[N], dp[N], sz[N], mx[N], mx2[N];
vector<int> edge[N]; void dfs1(int u, int fa, int val) {
sz[u] = ;
mx[u] = , mx2[u] = , dp[u] = ;
for(int v : edge[u]) {
if(v == fa) continue;
dfs1(v, u, val);
sz[u] += sz[v];
if(dp[v] == sz[v]) dp[u] += dp[v];
else {
if(dp[v] >= mx[u]) mx2[u] = mx[u], mx[u] = dp[v];
else if(dp[v] > mx2[u]) mx2[u] = dp[v];
}
}
dp[u] += mx[u];
if(a[u] < val) dp[u] = ;
}
void dfs2(int u, int fa, int cnt, int val, int &ans) {
if(!dp[u]) {
for(int v : edge[u]) {
if(v == fa) continue;
dfs2(v, u, , val, ans);
}
} else {
int ret = dp[u];
if(cnt == n - sz[u]) ret = max(ret, dp[u] + cnt);
else if(cnt > mx[u]) ret = max(ret, dp[u] - mx[u] + cnt), mx2[u] = mx[u], mx[u] = cnt;
else if(cnt > mx2[u]) mx2[u] = cnt;
ans = max(ans, ret);
for(int v : edge[u]) {
if(v == fa) continue;
if(dp[v] == sz[v]) dfs2(v, u, ret-dp[v], val, ans);
else if(dp[v] == mx[u]) dfs2(v, u, ret-mx[u]+mx2[u], val, ans);
else dfs2(v, u, ret, val, ans);
}
}
} bool check(int val) {
dfs1(, , val);
int ans = ;
dfs2(, , , val, ans);
return ans >= k;
} int main() {
scanf("%d%d", &n, &k);
for(int i = ; i <= n; i++) scanf("%d", &a[i]);
for(int i = ; i < n; i++) {
int u, v; scanf("%d%d", &u, &v);
edge[u].push_back(v);
edge[v].push_back(u);
}
int low = , high = , ans = ;
while(low <= high) {
int mid = low + high >> ;
if(check(mid)) ans = mid, low = mid + ;
else high = mid - ;
}
printf("%d\n", ans);
return ;
} /*
*/

8VC Venture Cup 2016 - Final Round (Div. 1 Edition) E - Preorder Test 树形dp的更多相关文章

  1. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition)

    暴力 A - Orchestra import java.io.*; import java.util.*; public class Main { public static void main(S ...

  2. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) A

    A. Orchestra time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  3. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) D. Factory Repairs 树状数组

    D. Factory Repairs 题目连接: http://www.codeforces.com/contest/635/problem/D Description A factory produ ...

  4. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) C. XOR Equation 数学

    C. XOR Equation 题目连接: http://www.codeforces.com/contest/635/problem/C Description Two positive integ ...

  5. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition)B. sland Puzzle 水题

    B. sland Puzzle 题目连接: http://www.codeforces.com/contest/635/problem/B Description A remote island ch ...

  6. 8VC Venture Cup 2016 - Final Round (Div. 2 Edition) A. Orchestra 水题

    A. Orchestra 题目连接: http://www.codeforces.com/contest/635/problem/A Description Paul is at the orches ...

  7. Codeforces Round #393 (Div. 2) (8VC Venture Cup 2017 - Final Round Div. 2 Edition) E - Nikita and stack 线段树好题

    http://codeforces.com/contest/760/problem/E 题目大意:现在对栈有m个操作,但是顺序是乱的,现在每输入一个操作要求你输出当前的栈顶, 注意,已有操作要按它们的 ...

  8. Codeforces Round #393 (Div. 2) (8VC Venture Cup 2017 - Final Round Div. 2 Edition) D - Travel Card

    D - Travel Card 思路:dp,类似于单调队列优化. 其实可以写的更简单... #include<bits/stdc++.h> #define LL long long #de ...

  9. Codeforces Round #393 (Div. 2) (8VC Venture Cup 2017 - Final Round Div. 2 Edition)A 水 B 二分 C并查集

    A. Petr and a calendar time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

随机推荐

  1. 关于ARGB_8888、ALPHA_8、ARGB_4444、RGB_565的理解

    关于ARGB_8888.ALPHA_8.ARGB_4444.RGB_565的理解 A:透明度 R:红色 G:绿 B:蓝 Bitmap.Config ARGB_4444:每个像素占四位,即A=4,R=4 ...

  2. bzoj千题计划122:bzoj1034: [ZJOI2008]泡泡堂BNB

    http://www.lydsy.com/JudgeOnline/problem.php?id=1034 从小到大排序后 最大得分: 1.自己最小的>对方最小的,赢一场 2.自己最大的>对 ...

  3. CF869 E 二维BIT

    1代表建一个屏障,2代表去掉一个屏障,3询问是否两点相通. 仿造一维询问是否在同一区间的问题扩展到二维,树状数组维护区间标记即可,标记值可以直接2500进制不会爆LL. /** @Date : 201 ...

  4. 【转】(总结)Nginx配置文件nginx.conf中文详解

    本文转载自:http://www.ha97.com/5194.html 定义Nginx运行的用户和用户组 user www www; nginx进程数,建议设置为等于CPU总核心数 worker_pr ...

  5. Java时间格式转换工具类

    把当前时间修改成指定时间 //把当前时间修改成指定时间 public String dateUtil(Integer seconds, String dateFormatPattern){ Date ...

  6. Linux IO模型

    1. Linux IO 模型矩阵 2. 同步阻塞IO 3. 同步非阻塞IO 4. 异步阻塞IO 5. 异步非阻塞IO

  7. HDU 1718 Rank 排序

    解题报告:给一个班的学生的分数排序,然后判断Jack在他们班级的排名是多少,并且有如下规定,若多个人的分数相同,则他们的排名也 是相同的.说白了就是问这个班上分数比Jack高的人数有多少个,如果有n个 ...

  8. linux tar 解压出错

    今天用tar -xzvf php-7.2.3.tar.gz 解压php的tar包时报错 [root@VM_72_37_centos ~]# tar -xzvf php-.tar.gz gzip: st ...

  9. 【译】第十篇 Integration Services:高级事件行为

    本篇文章是Integration Services系列的第十篇,详细内容请参考原文. 简介在前一篇, we introduced fault tolerance by examining method ...

  10. 可能是最漂亮的Spring事务管理详解

    Java面试通关手册(Java学习指南):https://github.com/Snailclimb/Java_Guide 微信阅读地址链接:可能是最漂亮的Spring事务管理详解 事务概念回顾 什么 ...