Codeforces 1082 C. Multi-Subject Competition-有点意思 (Educational Codeforces Round 55 (Rated for Div. 2))
2 seconds
256 megabytes
standard input
standard output
A multi-subject competition is coming! The competition has mm different subjects participants can choose from. That's why Alex (the coach) should form a competition delegation among his students.
He has nn candidates. For the ii-th person he knows subject sisi the candidate specializes in and riri — a skill level in his specialization (this level can be negative!).
The rules of the competition require each delegation to choose some subset of subjects they will participate in. The only restriction is that the number of students from the team participating in each of the chosen subjects should be the same.
Alex decided that each candidate would participate only in the subject he specializes in. Now Alex wonders whom he has to choose to maximize the total sum of skill levels of all delegates, or just skip the competition this year if every valid non-empty delegation has negative sum.
(Of course, Alex doesn't have any spare money so each delegate he chooses must participate in the competition).
The first line contains two integers nn and mm (1≤n≤1051≤n≤105, 1≤m≤1051≤m≤105) — the number of candidates and the number of subjects.
The next nn lines contains two integers per line: sisi and riri (1≤si≤m1≤si≤m, −104≤ri≤104−104≤ri≤104) — the subject of specialization and the skill level of the ii-th candidate.
Print the single integer — the maximum total sum of skills of delegates who form a valid delegation (according to rules above) or 00 if every valid non-empty delegation has negative sum.
6 3
2 6
3 6
2 5
3 5
1 9
3 1
22
5 3
2 6
3 6
2 5
3 5
1 11
23
5 2
1 -1
1 -5
2 -1
2 -1
1 -10
0
In the first example it's optimal to choose candidates 11, 22, 33, 44, so two of them specialize in the 22-nd subject and other two in the 33-rd. The total sum is 6+6+5+5=226+6+5+5=22.
In the second example it's optimal to choose candidates 11, 22 and 55. One person in each subject and the total sum is 6+6+11=236+6+11=23.
In the third example it's impossible to obtain a non-negative sum.
题意就是选科目,每科人数必须相同,总和尽量大。
有一个坑,可以往里面加入负数,只要该科总和>0就可以,具体代码。
代码:
//C
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+;
const int inf=0x3f3f3f3f; vector<int> ve[maxn];
vector<ll> sum[maxn];
int sz[maxn];
ll ans[maxn]; bool cmp(int a,int b)
{
return a>b;
} int main()
{
int n,m;
cin>>n>>m;
for(int i=;i<=n;i++){
int s,r;
cin>>s>>r;
ve[s].push_back(r);
}
for(int i=;i<=m;i++)
sort(ve[i].begin(),ve[i].end(),cmp);
for(int i=;i<=m;i++){
int pre=;
for(int j=;j<ve[i].size();j++){
if(j==) sum[i].push_back(ve[i][j]),pre=ve[i][j];
else sum[i].push_back(pre+ve[i][j]),pre=sum[i][j];
}
}
int maxx=;
for(int i=;i<=m;i++){
maxx=max(maxx,(int)ve[i].size());
for(int j=;j<ve[i].size();j++){
ans[j]=max(ans[j],ans[j]+sum[i][j]);
}
}
ll ret=;
for(int i=;i<maxx;i++)
ret=max(ret,ans[i]);
cout<<ret<<endl;
}
Codeforces 1082 C. Multi-Subject Competition-有点意思 (Educational Codeforces Round 55 (Rated for Div. 2))的更多相关文章
- Educational Codeforces Round 55 (Rated for Div. 2):C. Multi-Subject Competition
C. Multi-Subject Competition 题目链接:https://codeforces.com/contest/1082/problem/C 题意: 给出n个信息,每个信息包含专业编 ...
- Educational Codeforces Round 55 (Rated for Div. 2) C. Multi-Subject Competition 【vector 预处理优化】
传送门:http://codeforces.com/contest/1082/problem/C C. Multi-Subject Competition time limit per test 2 ...
- Educational Codeforces Round 55 (Rated for Div. 2) C. Multi-Subject Competition (实现,贪心,排序)
C. Multi-Subject Competition time limit per test2 seconds memory limit per test256 megabytes inputst ...
- [Educational Codeforces Round 55 (Rated for Div. 2)][C. Multi-Subject Competition]
https://codeforc.es/contest/1082/problem/C 题目大意:有m个类型,n个人,每个人有一个所属类型k和一个能力v,要求所选的类型的人个数相等并且使v总和最大(n, ...
- Codeforces 1082 D. Maximum Diameter Graph-树的直径-最长链-构造题 (Educational Codeforces Round 55 (Rated for Div. 2))
D. Maximum Diameter Graph time limit per test 2 seconds memory limit per test 256 megabytes input st ...
- Codeforces 1082 B. Vova and Trophies-有坑 (Educational Codeforces Round 55 (Rated for Div. 2))
B. Vova and Trophies time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- Codeforces 1082 A. Vasya and Book-题意 (Educational Codeforces Round 55 (Rated for Div. 2))
A. Vasya and Book time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- Educational Codeforces Round 55 (Rated for Div. 2) A/B/C/D
http://codeforces.com/contest/1082/problem/A WA数发,因为默认为x<y = = 分情况讨论,直达 or x->1->y or x-& ...
- Educational Codeforces Round 55 (Rated for Div. 2) B. Vova and Trophies 【贪心 】
传送门:http://codeforces.com/contest/1082/problem/B B. Vova and Trophies time limit per test 2 seconds ...
随机推荐
- SpringMVC 用注解Annotation驱动的IoC功能@Autowired @Component
转载自:http://blog.csdn.net/lufeng20/article/details/7598564 本文分为三个部分:概述.使用注解进行属性注入.使用注解进行Bean的自动定义. 一, ...
- ES6数组的扩展运算符
一.基本使用 ES6中函数可以使用 rest参数 接收函数的多余参数,组成一个数组,放在形参的最后面. let fn = (a, ...value) => { console.log(a); c ...
- 转【jenkins插件】
开源版本的Jenkins 具有三大能力:Master-Slave的分布式构建调度能力.Pipeline编排能力.强大的开源生态(插件)能力. 2017年4月,Jenkins创始人KK(Kohsuke ...
- Flask从入门到放弃1:路由app.route()
Flask从入门到放弃1: Flask中的路由app.route(): 参考来源:http://python.jobbole.com/80956/ https://www.raspberrypi.or ...
- 「6月雅礼集训 2017 Day8」gcd
[题目大意] 定义times(a, b)表示用辗转相除计算a和b的最大公约数所需步骤. 那么有: 1. times(a, b) = times(b, a) 2. times(a, 0) = 0 3. ...
- hdu 2157 How many ways?? ——矩阵十题第八题
Problem Description 春天到了, HDU校园里开满了花, 姹紫嫣红, 非常美丽. 葱头是个爱花的人, 看着校花校草竞相开放, 漫步校园, 心情也变得舒畅. 为了多看看这迷人的校园, ...
- 【NOI】2004 郁闷的出纳员
[算法]平衡树(treap) [题解] treap知识见数据结构. 解法,具体细节见程序. #include<cstdio> #include<algorithm> #incl ...
- hdu 1598 find the most comfortable road (并查集+枚举)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1598 find the most comfortable road Time Limit: 1000/ ...
- 树莓派开启smb
1.安装smb apt-get install samba samba-common-bin 2.修改/etc/samba/smb.conf配置 设置使用系统用户登入 增加smb访问文件夹 [shar ...
- 利用Python 发送邮件
概要 我们都知道SMTP(简单邮件传输协议),是一组用于从原地址到目的地址传输邮件的规范,通过它来控制邮件的中转方式.SMTP规定电子邮件应该如何格式化.如何加密,以及如何在邮件服务器之间传递.SMT ...