C. Multi-Subject Competitiontime

             C. Multi-Subject Competitiontime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputA 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).InputThe 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.OutputPrint 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.ExamplesinputCopy6 3

2 6
3 6
2 5
3 5
1 9
3 1
outputCopy22
inputCopy5 3
2 6
3 6
2 5
3 5
1 11
outputCopy23
inputCopy5 2
1 -1
1 -5
2 -1
2 -1
1 -10
outputCopy0
NoteIn 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.

#include<bits/stdc++.h>

#include<iostream>

#include<algorithm>

using namespace std;

#define ll long long

ll ans;ll l;ll k,c;ll maxx;ll fi,se;

pair<ll,ll> p[1000000];

ll a[1000000];

int main()

{

   
int m,n;

   
while(cin>>m>>n)

    {

       
maxx=0;

       
memset(a,0,sizeof(a));

       
for(int i=0;i<m;i++)

       
{

           
cin>>fi>>se;

           
p[i]={fi,-se};

       
}

       
sort(p,p+m);

       
for(int i=0;i<m;i++)

       
{

           
if(p[i].first!=l)

           
{

                k=0;c=0;

                l=p[i].first;

           
}

           
k++;

           
c-=p[i].second;

           
if(c>0)

           
a[k]+=c;

           
maxx=max(maxx,a[k]);

       
}

       
cout<<maxx<<endl;

    }

 

   
return 0;

}

动态规划，用于最大值或者最小值，依赖于前面一次的情况。
每一次累加，保存到数组，由于后面总体数量可能少，个体多。
pair的使用，排序的一些小技巧。
要用一个变量累加一类的，数组下标，关联。

• 注意清零
• 为什么做不出来呢，嗯，因为没看题目，要会灵活使用STL，变量关联。
  其实还是比较简单的，可惜脑袋抽了。