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#5727

Paired Up 1s 32MB

Problemas

There are a total of N (1≤N≤5000) cows on the number line, each of which is a Holstein or a Guernsey. The breed of the i-th cow is given by b_i∈{H,G}, the location of the i-th cow is given by x_i (0≤xi≤10^9), and the weight of the i-th cow is given by y_i (1≤y_i≤10^5).

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

  • Every pair consists of a Holstein h and a Guernsey g whose locations are within K of each other (1≤K≤10^9); that is, |x_h−x_g|≤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=1, compute the minimum possible sum of weights of the unpaired cows.

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


Entrada

The first input line contains T, N, and K.

Following this are N lines, the i-th of which contains b_i,x_i,y_i. It is guaranteed that 0≤x_1<x_2<⋯<x_N≤10^9.


Salida

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


Ejemplo #1

2 5 4
G 1 1
H 3 4
G 4 2
H 6 6
H 8 9
16

Cows 2 and 3 can pair up because they are at distance 1, which is at most K=4. This pairing is maximal, because cow 1, the only remaining Guernsey, is at distance 5 from cow 4 and distance 7 from cow 5, which are more than K=4. The sum of weights of unpaired cows is 1+6+9=16.


Ejemplo #2

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

Cows 1 and 2 can pair up because they are at distance 2≤K=4, and cows 3 and 5 can pair up because they are at distance 4≤K=4. This pairing is maximal because only cow 4 remains. The sum of weights of unpaired cows is the weight of the only unpaired cow, which is simply 6.


Ejemplo #3

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
1893

18+465+870+540=1893


Fuente

USACO 2021 December Platinum

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