codeforces # 317 div 2 B. Order Book（模拟）

B. Order Book

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In this task you need to process a set of stock exchange orders and use them to create order book.

An order is an instruction of some participant to buy or sell stocks on stock exchange. The order number i has price pi, direction di — buy or sell, and integer qi. This means that the participant is ready to buy or sell qi stocks at price pi for one stock. A value qi is also known as a volume of an order.

All orders with the same price p and direction d are merged into one aggregated order with price p and direction d. The volume of such order is a sum of volumes of the initial orders.

An order book is a list of aggregated orders, the first part of which contains sell orders sorted by price in descending order, the second contains buy orders also sorted by price in descending order.

An order book of depth s contains s best aggregated orders for each direction. A buy order is better if it has higher price and a sell order is better if it has lower price. If there are less than s aggregated orders for some direction then all of them will be in the final order book.

You are given n stock exhange orders. Your task is to print order book of depth s for these orders.

Input

The input starts with two positive integers n and s (1 ≤ n ≤ 1000, 1 ≤ s ≤ 50), the number of orders and the book depth.

Next n lines contains a letter di (either 'B' or 'S'), an integer pi (0 ≤ pi ≤ 105) and an integer qi (1 ≤ qi ≤ 104) — direction, price and volume respectively. The letter 'B' means buy, 'S' means sell. The price of any sell order is higher than the price of any buy order.

Output

Print no more than 2s lines with aggregated orders from order book of depth s. The output format for orders should be the same as in input.

Sample test(s)

input

6 2
B 10 3
S 50 2
S 40 1
S 50 6
B 20 4
B 25 10

output

S 50 8
S 40 1
B 25 10
B 20 4

Note

Denote (x, y) an order with price x and volume y. There are 3 aggregated buy orders (10, 3), (20, 4), (25, 10) and two sell orders (50, 8), (40, 1) in the sample.

You need to print no more than two best orders for each direction, so you shouldn't print the order (10 3) having the worst price among buy orders.

昨天头疼。。然后觉得要掉。。就用小号打的。。。

结果涨了２００＋rating。。。。

这题就是纯模拟啊。。。

一遍ac。。。

需要注意的地方，如果说有。。。

大概就是，sell的是取最小的s个，要先从小到大排序。。。但是输出的时候又是从大到小排序。。。

昨天４０分钟做完前2道题。。然后c题不会搞。。。就水群（求不鄙视> <），好像一堆人wa　b　wa得特别惨的样子。。。

/*************************************************************************
    > File Name: code/cf/#317/B.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月23日 星期日 00时30分51秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define ms(a,x) memset(a,x,sizeof(a))
#define lr dying111qqz
using namespace std;
#define For(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef double DB;
const int inf = 0x3f3f3f3f;
const int N=1E5+7;
struct Q
{
    int val;
    int num;
    int id;
}b[N],s[N],ans[N];
int id[N];
int id2[N];
int n,ss;

bool cmp(Q a,Q b)
{
    if (a.val<b.val)
    return true;
    return false;
}
bool cmp2(Q a,Q b)
{
    if (a.val>b.val) return true;
    return false;
}
int main()
{
  #ifndef  ONLINE_JUDGE 
    freopen("in.txt","r",stdin); 
  #endif
    cin>>n>>ss;
    int x,y;
    char order;
    ms(b,0);
    ms(s,0);
    ms(id,0);
    ms(id2,0);
    int cntA=0,cntB= 0 ;
    for ( int i  = 0  ; i < n ; i++){
    cin>>order>>x>>y;
    if (order=='B'){
        if (id[x]==0)
        {
        cntA++;
        id[x] = cntA;
        }
        b[id[x]].num = b[id[x]].num + y;
        b[id[x]].val = x;
        b[id[x]].id = id[x];
    }
    else
    {
        if (id2[x]==0)
        {
        cntB++;
        id2[x] = cntB;
        }
        s[id2[x]].num = s[id2[x]].num+y;
        s[id2[x]].val = x;
        s[id2[x]].id = id2[x];
    }
    }
    sort(s+1,s+cntB+1,cmp);
    sort(b+1,b+cntA+1,cmp2);
    cntA = min(cntA,ss);
    cntB = min(cntB,ss);
    ms(ans,0);
    for ( int i = 1 ; i <= cntB ; i++){
    ans [i].val = s[i].val;
    ans[i].num = s[i].num;
    }
    sort(ans+1,ans+cntB+1,cmp2);
    for ( int i = 1; i <= cntB ; i++){
    cout<<"S "<<ans[i].val<<" "<<ans[i].num<<endl;
    }
    ms(ans,0);
    for ( int i =  1 ; i <= cntA ; i++){
    ans[i].val = b[i].val;
    ans[i].num = b[i].num;
    }
    sort(ans+1,ans+cntA+1,cmp2);
    for ( int i = 1 ; i <= cntA ; i++){
    cout<<"B "<<ans[i].val<<" "<<ans[i].num<<endl;
    }
  
  
 #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}