#5159. Problem 2. Paired Up

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Problem 2. Paired Up

Problem 2. Paired Up

USACO 2021 December Contest, Platinum

There are a total of NN (1N50001\le N\le 5000) cows on the number line, each of which is a Holstein or a Guernsey. The breed of the ii-th cow is given by bi{H,G}b_i\in \{H,G\}, the location of the ii-th cow is given by xix_i (0xi1090 \leq x_i \leq 10^9), and the weight of the ii-th cow is given by yiy_i (1yi1051 \leq y_i \leq 10^5).

At Farmer John's signal, some of the cows will form pairs such that

  • Every pair consists of a Holstein hh and a Guernsey gg whose locations are within KK of each other (1K1091\le K\le 10^9); that is, xhxgK|x_h-x_g|\le K.
  • Every cow is either part of a single pair or not part of a pair.
  • The pairing is maximal; that is, no two unpaired cows can form a pair.

It's up to you to determine the range of possible sums of weights of the unpaired cows. Specifically,

  • If T=1T=1, compute the minimum possible sum of weights of the unpaired cows.
  • If T=2T=2, compute the maximum possible sum of weights of the unpaired cows.

INPUT FORMAT (input arrives from the terminal / stdin):

The first input line contains TT, NN, and KK.

Following this are NN lines, the ii-th of which contains bi,xi,yib_i,x_i,y_i. It is guaranteed that 0x1<x2<<xN1090\le x_1< x_2< \cdots< x_N\le 10^9.

OUTPUT FORMAT (print output to the terminal / stdout):

The minimum or maximum possible sum of weights of the unpaired cows.

SAMPLE INPUT:


2 5 4
G 1 1
H 3 4
G 4 2
H 6 6
H 8 9

SAMPLE OUTPUT:


16

Cows 22 and 33 can pair up because they are at distance 11, which is at most K=4K = 4. This pairing is maximal, because cow 11, the only remaining Guernsey, is at distance 55 from cow 44 and distance 77 from cow 55, which are more than K=4K = 4. The sum of weights of unpaired cows is 1+6+9=161 + 6 + 9 = 16.

SAMPLE INPUT:


1 5 4
G 1 1
H 3 4
G 4 2
H 6 6
H 8 9

SAMPLE OUTPUT:


6

Cows 11 and 22 can pair up because they are at distance 2K=42 \leq K = 4, and cows 33 and 55 can pair up because they are at distance 4K=44 \leq K = 4. This pairing is maximal because only cow 44 remains. The sum of weights of unpaired cows is the weight of the only unpaired cow, which is simply 66.

SAMPLE INPUT:


2 10 76
H 1 18
H 18 465
H 25 278
H 30 291
H 36 202
G 45 96
G 60 375
G 93 941
G 96 870
G 98 540

SAMPLE OUTPUT:


1893

The answer to this example is 18+465+870+540=189318+465+870+540=1893.

SCORING: Test cases 4-7 satisfy T=1T=1.Test cases 8-14 satisfy T=2T=2 and N300N\le 300.Test cases 15-22 satisfy T=2T=2.

Note: the memory limit for this problem is 512MB, twice the default.

Problem credits: Benjamin Qi