Paired Up
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Contribute a translation- Time limit
- 2000 ms
- Memory limit
- 256 MB
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Statement
There are a total of $N$ ($1\le N\le 5000$) cows on the number line, each of which is a Holstein or a Guernsey. The breed of the $i$-th cow is given by $b_i\in \{H,G\}$, the location of the $i$-th cow is given by $x_i$ ($0 \leq x_i \leq 10^9$), and the weight of the $i$-th cow is given by $y_i$ ($1 \leq y_i \leq 10^5$).
At Farmer John's signal, some of the cows will form pairs such that
- Every pair consists of a Holstein $h$ and a Guernsey $g$ whose locations are within $K$ of each other ($1\le K\le 10^9$); that is, $|x_h-x_g|\le K$.
- 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 $T=1$, 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.
Input
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 $0\le x_1< x_2< \cdots< x_N\le 10^9$.
Scoring
- Test cases 4-7 satisfy $T=1$.
- Test cases 8-14 satisfy $T=2$ and $N\le 300$.
- Test cases 15-22 satisfy $T=2$.
Note: the memory limit for this problem is 512MB, twice the default.
Output
The minimum or maximum possible sum of weights of the unpaired cows.
Examples
2 5 4 G 1 1 H 3 4 G 4 2 H 6 6 H 8 9
16
1 5 4 G 1 1 H 3 4 G 4 2 H 6 6 H 8 9
6
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
Notes
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 $1 + 6 + 9 = 16$.
Cows $1$ and $2$ can pair up because they are at distance $2 \leq K = 4$, and cows $3$ and $5$ can pair up because they are at distance $4 \leq K = 4$. 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 $6$.
The answer to this example is $18+465+870+540=1893$.
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