簡單的排序 也可以通過雜湊完成

#include<iostream>
#include<algorithm>
#include<stdio.h>
#define rep(i,a,b) for(int i = a ; i < b ; i++)
#define rrep(i,a,b) for(int i = b-1 ; i >= a ; i--)
using namespace std;

struct node{
     int p , q;
}B[1000],S[1000],NOW,ANSS[1000];

bool cmpb(node a , node b)
{
     if(a.p == b.p)
          return a.q < b.q;
     return a.p > b.p;
}

bool cmps(node a , node b)
{
     if(a.p == b.p)
          return a.q < b.q;
     return a.p < b.p;
}

int main()
{
     int n , s , cntb = 0 , cnts = 0 ,cntanss = 0;
     char str[5];
     cin>> n >> s;
     rep(i,0,n)
     {
          scanf("%s%d%d",str,&NOW.p,&NOW.q);
          if(str[0] == 'B')     B[cntb++] = NOW;
          else S[cnts++] = NOW;
     }
     sort(B,B+cntb,cmpb);
     sort(S,S+cnts,cmps);
     int ss = s;
     rep(i,0,cnts)
     {
          if(i+1 < cnts &&S[i].p == S[i+1].p)
          {
               S[i+1].q +=S[i].q;
               continue;
          }
          ss--;
          ANSS[cntanss++] = S[i];
          if(ss == 0) break;
     }
     sort(ANSS,ANSS+cntanss,cmpb);
     rep(i,0,cntanss) printf("S %d %d\n",ANSS[i].p,ANSS[i].q);
     rep(i,0,cntb)
     {
          if(i+1 < cntb &&B[i].p == B[i+1].p)
          {
               B[i+1].q +=B[i].q;
               continue;
          }
          s--;
          printf("B %d %d\n",B[i].p,B[i].q);
          if(s == 0) break;
     }
}