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

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.

跪在英文题意。。

题意： 对n个订购者，对卖家按降序输出小单价的s组结果；对买家按降序输出大单价的s组结果。相同单价的合并，若分别少于s组，则尽可能多输出。

思路：能看懂题意应该都能实现。

赛码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
#define CLR(arr,x) memset(arr,x,sizeof(arr))
using namespace std ;
const int maxn = 100000+10 ;
const int inf = 1<<30 ;
int init1[maxn] ,init2[maxn] ;
struct pm{
int p ,q ;
};
pm ss[1100] , b[1100] ;
bool cmp(pm a, pm b) {
if( a.p == b.p ) return a.q>b.q ;
return a.p > b.p ;
}
int main() {
int n , s ;
while(~scanf("%d%d%*c",&n,&s)){
CLR(init1,0) ;
CLR(init2,0) ;
char ch ;
int temp1 , temp2 ;
int num1 , num2 ;
num1 = num2 = 0 ;
for( int i = 0 ; i < n ; ++i ) {
scanf("%c",&ch) ;
if( ch == 'B' ) {
scanf("%d %d",&temp1,&temp2)  ;
init1[temp1] +=temp2 ;
ss[num1].p = temp1 ;
ss[num1].q = init1[temp1] ;
num1++;
}
else {
scanf("%d %d",&temp1,&temp2)  ;
init2[temp1] +=temp2 ;
b[num2].p = temp1 ;
b[num2].q = init2[temp1] ;
num2++;
}
getchar() ;
}

ss[num1].p = ss[num1].q = -1 ;
sort(b,b+num2,cmp) ;
sort(ss,ss+num1,cmp) ;
int temp3 = -1 , numm = 0,numm1 = 0 ,wz;
for(  wz = num2-1; wz >=0&&numm1<s+1 ; --wz){
if( b[wz].p != temp3 ) { temp3 = b[wz].p ; numm1++; }
}
if(wz == -1 && (numm1 != (s+1)) ) wz = 0 ;
else if(numm1 == s+1 ) {wz = wz+2 ;}
temp3 = -1 ;
for( int i = wz ; i < num2  && numm < s; ++i ) {
if( (b[i].p!=temp3)&&(b[i].p!=-1) ){
printf("S %d %d\n",b[i].p,b[i].q) ;
temp3 = b[i].p ;
numm++;
}
}
temp3 = -1 , numm = 0 ;
for( int i = 0 ; i < num1  && numm < s; ++i ) {
if( (ss[i].p!=temp3)&&(ss[i].p!=-1) ){
printf("B %d %d\n",ss[i].p,ss[i].q) ;
temp3 = ss[i].p ;
numm++;
}
}

}
return 0 ;
}