B. Order Book------（Codeforces Round #317 [AimFund Thanks-Round] (Div. 2)）

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本文详细探讨了如何处理股票交易所订单并构建订单簿，包括订单定义、订单合并及深度为s的订单簿生成过程。

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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.

分析：题目量有点大，一开始没有仔细看题，这是一个伤，导致我wa了很多次，后来AC了，1000分的题拿了360，最后的结果是被hack了，审题很重要。

CODE：

#include <bits/stdc++.h>
using namespace std;

const int maxn=100005;

struct node{
    int num;
    int pri;
}sp[100005],bp[100005];

int cmp(node a,node b){
    if(a.pri!=0&&b.pri!=0)
        return a.pri<b.pri;
    return a.num<b.num;
}

int main()
{
    int n,s;
    while(cin>>n>>s){
        char c;
        int a,b;
        memset(sp,0,sizeof(sp));
        memset(bp,0,sizeof(bp));
        for(int i=0;i<n;i++){
            cin>>c>>a>>b;
            if(c=='S'){
                sp[a].num+=b;
                sp[a].pri=a;
            }
            else{
                bp[a].num+=b;
                bp[a].pri=a;
            }
        }
        sort(bp,bp+maxn,cmp);
        sort(sp,sp+maxn,cmp);
        int cnt=0;
        for(int i=maxn-1;i>=0;i--){
            if(sp[i].num)
                cnt++;
            else
                break;
        }
        if(cnt>=s){
            for(int i=maxn-cnt+s-1;i>=maxn-cnt;i--){
                cout<<"S "<<sp[i].pri<<' '<<sp[i].num<<endl;
            }
        }
        else{
            for(int i=maxn-1;i>=0;i--){
                if(sp[i].num)
                    cout<<"S "<<sp[i].pri<<' '<<sp[i].num<<endl;
                else
                    break;
            }
        }
        for(int i=maxn-1;i>=maxn-s;i--){
            if(bp[i].num)
                cout<<"B "<<bp[i].pri<<' '<<bp[i].num<<endl;
            else
                break;
        }
    }
    return 0;
}