E - Aquarium decoration

枚举两个人都喜欢的个数,就能得到单个喜欢的个数,然后用平衡树维护前k大的和。

#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 = ;
const int Mod = 1e9 + ; int n, m, k, num, cnt, state[N], d[N], a[N], b[N], c[N];
LL suma[N], sumb[N], sumc[N]; struct node {
node* ch[];
int key, fix, sz, cnt;
LL sum;
void update() {
sz = ch[]->sz + ch[]->sz + cnt;
sum = ch[]->sum + ch[]->sum + 1ll*cnt*key;
}
}; typedef node* P_node; struct Treap {
node base[N], nil;
P_node root, null, len;
Treap() {
root = null = &nil;
null->key = null->fix = 1e9;
null->sz = null->cnt = ;
null->ch[] = null->ch[] = null;
len = base;
}
P_node newnode(int tkey) {
len->key = tkey;
len->fix = rand();
len->ch[] = len->ch[] = null;
len->sz = len->cnt = ;
len->sum = tkey;
return len++;
}
void rot(P_node &p, int d) {
P_node k = p->ch[d ^ ];
p->ch[d ^ ] = k->ch[d];
k->ch[d] = p;
p->update();
k->update();
p = k;
}
void _Insert(P_node &p, int tkey) {
if(p == null) {
p = newnode(tkey);
} else if(p->key == tkey) {
p->cnt++;
} else {
int d = tkey > p->key;
_Insert(p->ch[d], tkey);
if(p->ch[d]->fix > p->fix) {
rot(p, d ^ );
}
}
p->update();
} void _Delete(P_node &p, int tkey) {
if(p == null) return;
if(p->key == tkey) {
if(p->cnt > ) p->cnt--;
else if(p->ch[] == null) p = p->ch[];
else if(p->ch[] == null) p = p->ch[];
else {
int d = p->ch[]->fix > p->ch[]->fix;
rot(p, d);
_Delete(p->ch[d], tkey);
}
} else {
_Delete(p->ch[tkey > p->key], tkey);
}
p->update();
}
int _Kth(P_node p, int k) {
if(p == null || k < || k > p->sz) return ;
if(k < p->ch[]->sz + ) return _Kth(p->ch[], k);
if(k > p->ch[]->sz + p->cnt) return _Kth(p->ch[], k - p->ch[]->sz - p->cnt);
return p->key;
}
int _Rank(P_node p, int tkey, int res) {
if(p == null) return -;
if(p->key == tkey) return p->ch[]->sz + res + ;
if(tkey < p->key) return _Rank(p->ch[], tkey, res);
return _Rank(p->ch[], tkey, res + p->ch[]->sz + p->cnt);
}
int _Pred(P_node p, int tkey){
if(p == null) return -1e9;
if(tkey <= p->key) return _Pred(p->ch[], tkey);
return max(p->key, _Pred(p->ch[], tkey));
}
int _Succ(P_node p, int tkey){
if(p == null) return 1e9;
if(tkey >= p->key) return _Succ(p->ch[], tkey);
return min(p->key, _Succ(p->ch[], tkey));
}
LL _Query(P_node p, int res) {
if(!res) return ;
if(p->ch[]->sz >= res) return _Query(p->ch[], res);
else if(p->ch[]->sz + p->cnt < res) {
return p->ch[]->sum + 1ll*p->key*p->cnt + _Query(p->ch[], res - p->ch[]->sz - p->cnt);
} else {
return p->ch[]->sum + 1ll*p->key*(res - p->ch[]->sz);
}
}
void Insert(int tkey){ _Insert(root,tkey); }
void Delete(int tkey){ _Delete(root,tkey); }
int Kth(int k){ return _Kth(root,k); }
int Rank(int tkey){ return _Rank(root,tkey,); }
int Pred(int tkey){ return _Pred(root,tkey); }
int Succ(int tkey){ return _Succ(root,tkey); }
LL Query(int res){ return _Query(root,res); }
}tp; int main() {
scanf("%d%d%d", &n, &m, &k);
for(int i = ; i <= n; i++) scanf("%d", &d[i]);
scanf("%d", &num);
for(int i = ; i <= num; i++) {
int x; scanf("%d", &x);
state[x] |= ;
}
scanf("%d", &num);
for(int i = ; i <= num; i++) {
int x; scanf("%d", &x);
state[x] |= ;
}
for(int i = ; i <= n; i++) {
if(state[i] == ) tp.Insert(d[i]), cnt++;
else if(state[i] == ) a[++a[]] = d[i];
else if(state[i] == ) b[++b[]] = d[i];
else c[++c[]] = d[i];
}
sort(a + , a + + a[]);
sort(b + , b + + b[]);
sort(c + , c + + c[]);
for(int i = ; i <= a[]; i++)
suma[i] = suma[i-] + a[i];
for(int i = ; i <= b[]; i++)
sumb[i] = sumb[i-] + b[i];
for(int i = ; i <= c[]; i++)
sumc[i] = sumc[i-] + c[i]; LL ans = INF;
for(int i = ; i <= c[]; i++) tp.Insert(c[i]), cnt++;
for(int i = ; i <= c[]; i++) {
if(i) tp.Delete(c[i]), cnt--;
int res1 = max(, k - i);
if(a[] < res1 || b[] < res1) continue;
while(a[] > res1) {
tp.Insert(a[a[]]);
a[]--; cnt++;
}
while(b[] > res1) {
tp.Insert(b[b[]]);
b[]--; cnt++;
}
if(i + * res1 > m) continue;
int res2 = m - i - * res1;
ans = min(ans, sumc[i] + suma[a[]] + sumb[b[]] + tp.Query(res2));
}
printf("%lld\n", ans == INF ? - : ans);
return ;
} /*
*/

Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) E - Aquarium decoration 贪心 + 平衡树的更多相关文章

  1. Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)(A.暴力,B.优先队列,C.dp乱搞)

    A. Carrot Cakes time limit per test:1 second memory limit per test:256 megabytes input:standard inpu ...

  2. C.Fountains(Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)+线段树+RMQ)

    题目链接:http://codeforces.com/contest/799/problem/C 题目: 题意: 给你n种喷泉的价格和漂亮值,这n种喷泉题目指定用钻石或现金支付(分别用D和C表示),C ...

  3. Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) C. Fountains 【树状数组维护区间最大值】

    题目传送门:http://codeforces.com/contest/799/problem/C C. Fountains time limit per test 2 seconds memory ...

  4. Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) 一夜回到小学生

    我从来没想过自己可以被支配的这么惨,大神讲这个场不容易掉分的啊 A. Carrot Cakes time limit per test 1 second memory limit per test 2 ...

  5. 【动态规划】【滚动数组】【搜索】Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) D. Field expansion

    显然将扩张按从大到小排序之后,只有不超过前34个有效. d[i][j]表示使用前i个扩张,当length为j时,所能得到的最大的width是多少. 然后用二重循环更新即可, d[i][j*A[i]]= ...

  6. 【预处理】【分类讨论】Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) C. Fountains

    分几种情况讨论: (1)仅用C或D买两个 ①买两个代价相同的(实际不同)(排个序) ②买两个代价不同的(因为买两个代价相同的情况已经考虑过了,所以此时对于同一个代价,只需要保存美丽度最高的喷泉即可)( ...

  7. 树状数组 Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) C. Fountains

    C. Fountains time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  8. Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) D. Field expansion

    D. Field expansion time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  9. Codeforces Round #413, rated, Div. 1 + Div. 2 C. Fountains(贪心 or 树状数组)

    http://codeforces.com/contest/799/problem/C 题意: 有n做花园,有人有c个硬币,d个钻石 (2 ≤ n ≤ 100 000, 0 ≤ c, d ≤ 100  ...

随机推荐

  1. javamail模拟邮箱功能获取邮件内容-中级实战篇【内容|附件下载方法】(javamail API电子邮件实例)

    引言: JavaMail jar包下载地址:http://java.sun.com/products/javamail/downloads/index.html 此篇是紧随上篇文章而封装出来的,阅读本 ...

  2. An Insight to References in C++

    [An Insight to References in C++] 引用的本质是常指针.占用的内存和指针一样. 参考:http://www.codeproject.com/Articles/13363 ...

  3. AngularJS 、Backbone.js 和 Ember.js 的比较

    1 介绍 我们准备在这篇文章中比较三款流行于Web的“模型-视图-*”框架:AngularJS.Backbone和Ember.为你的项目选择正确的框架能够对你及时交付项目的能力和在以后维护你自己代码的 ...

  4. 如何在python的字符串中输入纯粹的{}

    python的format函数通过{}来格式化字符串 >>> a='{0}'.format(123) >>> a ' 如果需要在文本中包含{}字符,这样使用就会报错 ...

  5. 原生的js实现jsonp的跨域封装

    一.原理 jsonp是利用浏览器请求script文件时不受同源策略的限制而实现的,伪造一个script标签,将请求数据的url赋值给script的src属性,并将该标签添加到html中,浏览器会自动发 ...

  6. Shell-help格式详解

    前言 linux shell命令通常可以通过-h或--help来打印帮助说明,或者通过man命令来查看帮助,有时候我们也会给自己的程序写简单的帮助说明,其实帮助说明格式是有规律可循的 帮助示例 下面是 ...

  7. 从此编写 Bash 脚本不再难【转】

    从此编写 Bash 脚本不再难 原创 Linux技术 2017-05-02 14:30 在这篇文章中,我们会介绍如何通过使用 bash-support vim 插件将 Vim 编辑器安装和配置 为一个 ...

  8. Prepare tasks for django project deployment.md

    As we know, there are some boring tasks while deploy Django project, like create db, do migrations a ...

  9. 25 The Go image/draw package go图片/描绘包:图片/描绘包的基本原理

    The Go image/draw package  go图片/描绘包:图片/描绘包的基本原理 29 September 2011 Introduction Package image/draw de ...

  10. vue全面介绍--全家桶、项目实例

    简介 “简单却不失优雅,小巧而不乏大匠”. 2016年最火的前端框架当属Vue.js了,很多使用过vue的程序员这样评价它,“vue.js兼具angular.js和react.js的优点,并剔除了它们 ...