AtCoder Beginner Contest 116 D - Various Sushi (贪心+栈)
D - Various Sushi
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 400400 points
Problem Statement
There are NN pieces of sushi. Each piece has two parameters: "kind of topping" titi and "deliciousness" didi. You are choosing KK among these NN pieces to eat. Your "satisfaction" here will be calculated as follows:
- The satisfaction is the sum of the "base total deliciousness" and the "variety bonus".
- The base total deliciousness is the sum of the deliciousness of the pieces you eat.
- The variety bonus is x∗xx∗x, where xx is the number of different kinds of toppings of the pieces you eat.
You want to have as much satisfaction as possible. Find this maximum satisfaction.
Constraints
- 1≤K≤N≤1051≤K≤N≤105
- 1≤ti≤N1≤ti≤N
- 1≤di≤1091≤di≤109
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
NN KK
t1t1 d1d1
t2t2 d2d2
..
..
..
tNtN dNdN
Output
Print the maximum satisfaction that you can obtain.
Sample Input 1 Copy
5 3
1 9
1 7
2 6
2 5
3 1
Sample Output 1 Copy
26
If you eat Sushi 1,21,2 and 33:
- The base total deliciousness is 9+7+6=229+7+6=22.
- The variety bonus is 2∗2=42∗2=4.
Thus, your satisfaction will be 2626, which is optimal.
Sample Input 2 Copy
7 4
1 1
2 1
3 1
4 6
4 5
4 5
4 5
Sample Output 2 Copy
25
It is optimal to eat Sushi 1,2,31,2,3 and 44.
Sample Input 3 Copy
6 5
5 1000000000
2 990000000
3 980000000
6 970000000
6 960000000
4 950000000
Sample Output 3 Copy
4900000016
Note that the output may not fit into a 3232-bit integer type.
题意:
给定N个结构体,每一个结构体有两个信息,分别是type 和 x,让你从中选出K个结构体,
使之type的类型数的平方+sum{ xi } 最大。
思路:
可以贪心来做。
首先根据x由大到小来排序,然后选入结构体,先贪心的全前K个,把每一种类型的第一个一定的加入到我们选择的范围中,(贪心思想)
然后把某一种类型的后面几个的结构体加入到栈中,
然后扫k+1~n的时候,如果这个结构体的类型没加入过,那么把他加入,然后类型数目+1,弹出栈顶的元素,减去他的x值贡献,然后用新结构体取尝试更新最大值。
还主要用到了stack的先进后出的思想,巧妙的最优的替换了每一个能替换掉的是当前选择中贡献最小的。
贪心好题,(口胡结束)。细节见代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rt return
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n,k;
struct info
{
int x,v;
}a[maxn];
bool cmp(info one,info two)
{
return one.v>two.v;
}
int vis[maxn];
stack<int> s;
int main()
{
scanf("%d %d",&n,&k);
repd(i,,n)
{
scanf("%d %d",&a[i].x,&a[i].v);
}
sort(a+,a++n,cmp);
ll cnt=;
ll tp=;
ll res=;
ll ans=0ll;
repd(i,,n)
{
if(cnt<k)
{
if(vis[a[i].x]==)
{
vis[a[i].x]=;
tp++;
}else
{
s.push(a[i].v);
}
res+=a[i].v;
cnt++;
ans=max(ans,res+1ll*tp*tp);
}
else{
if(s.empty())
break;
if(vis[a[i].x])
continue;
vis[a[i].x]=;
tp++;
res-=s.top();
res+=a[i].v;
s.pop();
ans=max(ans,res+tp*tp);
}
}
printf("%lld\n",ans);
return ;
} inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '');
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * - ch + '';
}
}
else {
*p = ch - '';
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * + ch - '';
}
}
}
AtCoder Beginner Contest 116 D - Various Sushi (贪心+栈)的更多相关文章
- AtCoder Beginner Contest 116 D - Various Sushi 【贪心+栈】
Problem Statement There are NN pieces of sushi. Each piece has two parameters: "kind of topping ...
- Atcoder Beginner Contest 118 C-Monsters Battle Royale(贪心)
题目链接 题意就是要让给出的数字去互相取余,看看能得到最小的数事多少. 那么就可以从小到大排序,每一次都贪心地把最小的数作为攻击者,去攻击其他的数字(也就是大的取余小的),然后再一次排序,循环这个过程 ...
- AtCoder Beginner Contest 137 D题【贪心】
[题意]一共有N个任务和M天,一个人一天只能做一个任务,做完任务之后可以在这一天之后的(Ai-1)天拿到Bi的工资,问M天内最多可以拿到多少工资. 链接:https://atcoder.jp/cont ...
- AtCoder Beginner Contest 181 E - Transformable Teacher (贪心,二分)
题意:有一长度为奇数\(n\)的数组\(a\),和长度为\(m\)的数组\(b\),现要求从\(b\)中选择一个数放到\(a\)中,并将\(a\)分成\((n+1)/2\)个数对,求最小的所有数对差的 ...
- AtCoder Beginner Contest 249 F - Ignore Operations // 贪心 + 大根堆
传送门:F - Keep Connect (atcoder.jp) 题意: 给定长度为N的操作(ti,yi). 给定初值为0的x,对其进行操作:当t为1时,将x替换为y:当t为2时,将x加上y. 最多 ...
- AtCoder Beginner Contest 116 C题 【题意:可以在任意区间【L,R】上加1,求通过最少加1次数得到题目给定的区间】】{思维好题}
C - Grand Garden In a flower bed, there are NN flowers, numbered 1,2,......,N1,2,......,N. Initially ...
- AtCoder Beginner Contest 153 题解
目录 AtCoder Beginner Contest 153 题解 A - Serval vs Monster 题意 做法 程序 B - Common Raccoon vs Monster 题意 做 ...
- AtCoder Beginner Contest 173 题解
AtCoder Beginner Contest 173 题解 目录 AtCoder Beginner Contest 173 题解 A - Payment B - Judge Status Summ ...
- AtCoder Beginner Contest 172 题解
AtCoder Beginner Contest 172 题解 目录 AtCoder Beginner Contest 172 题解 A - Calc B - Minor Change C - Tsu ...
随机推荐
- Sql Server XML
实验数据: Create table xmldata (name NVARCHAR(20), age int, sex NVARCHAR(5) ) INSERT INTO xmldata VALUES ...
- MySQL各类SQL语句的加锁机制
官网参考:https://dev.mysql.com/doc/refman/5.6/en/innodb-locks-set.html MySQL把读操作分为两大类:锁定读和非锁定读(即locking ...
- strlen strcat strcpy strcmp 自己实现
strlen strcat strcpy strcmp 自己实现 strlen include <stdio.h> #include <string.h> #include & ...
- Python - 判断list是否为空
Python中判断list是否为空有以下两种方式: 方式一: list_temp = [] if len(list_temp): # 存在值即为真 else: # list_temp是空的 方式二: ...
- SpringBoot实现热部署(修改class不需要重启)
热部署: devtools可以实现页面热部署(即页面修改后会立即生效, 这个可以直接在application.properties文件中配置spring.thymeleaf.cache=false来实 ...
- LeetCode算法题-Power of Four(Java实现-六种解法)
这是悦乐书的第205次更新,第216篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第72题(顺位题号是342).给定一个整数(带符号的32位),写一个函数来检查它是否为4 ...
- Python爬虫-01:爬虫的概念及分类
目录 # 1. 为什么要爬虫? 2. 什么是爬虫? 3. 爬虫如何抓取网页数据? # 4. Python爬虫的优势? 5. 学习路线 6. 爬虫的分类 6.1 通用爬虫: 6.2 聚焦爬虫: # 1. ...
- 03.Python网络爬虫第一弹《Python网络爬虫相关基础概念》
爬虫介绍 引入 之前在授课过程中,好多同学都问过我这样的一个问题:为什么要学习爬虫,学习爬虫能够为我们以后的发展带来那些好处?其实学习爬虫的原因和为我们以后发展带来的好处都是显而易见的,无论是从实际的 ...
- 13.scrapy框架的日志等级和请求传参
今日概要 日志等级 请求传参 如何提高scrapy的爬取效率 今日详情 一.Scrapy的日志等级 - 在使用scrapy crawl spiderFileName运行程序时,在终端里打印输出的就是s ...
- Mysql的用户管理