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

#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;

}


 

動態規劃，用於最大值或者最小值，依賴於前面一次的情況。
每一次累加，儲存到陣列，由於後面總體數量可能少，個體多。
pair的使用，排序的一些小技巧。
要用一個變數累加一類的，陣列下標，關聯。

• 注意清零
• 為什麼做不出來呢，嗯，因為沒看題目，要會靈活使用STL，變數關聯。
  其實還是比較簡單的，可惜腦袋抽了。