パナソニックプログラミングコンテスト(AtCoder Beginner Contest 195)
D
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⌀ 945
原题
← CE →
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Problem Statement

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We have $N$ pieces of baggage called Baggage $1$ through $N$, and $M$ boxes called Box $1$ through $M$.

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Baggage $i$ has a size of $W_i$ and a value of $V_i$.

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Box $i$ can contain a piece of baggage whose size of at most $X_i$. It cannot contain two or more pieces of baggage.

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You will be given $Q$ queries. In each query, given two integers $L$ and $R$, solve the following problem:

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  • Problem: Out of the $M$ boxes, $R-L+1$ boxes, Box $L,L+1,\ldots,R$, have become unavailable. Find the maximum possible total value of a set of baggage that we can put into the remaining boxes simultaneously.
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Constraints

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  • $1 \leq N \leq 50$
  • $1 \leq M \leq 50$
  • $1 \leq Q \leq 50$
  • $1 \leq W_i \leq 10^6$
  • $1 \leq V_i \leq 10^6$
  • $1 \leq X_i \leq 10^6$
  • $1 \leq L \leq R \leq M$
  • All values in input are integers.
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Input

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Input is given from Standard Input in the following format:

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N M Q
W_1 V_1
\vdots
W_N V_N
X_1 \ldots X_M
\mathrm{Query}_1
\vdots
\mathrm{Query}_Q

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Each Query is in the following format:

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L R

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Output

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Print $Q$ lines.

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The $i$-th line should contain the answer to the problem described by $\mathrm{Query}_i$.

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样例

Input #1
3 4 3
1 9
5 3
7 8
1 8 6 9
4 4
1 4
1 3
Output #1
20
0
9
In the $1$-st query, only Box $4$ is unavailable. By putting Baggage $1$ into Box $1$, Baggage $3$ into Box $2$, and Baggage $2$ into Box $3$, we can put all baggage into boxes, making the total value of baggage in boxes $20$. In the $2$-nd query, all boxes are unavailable; the answer is $0$. In the $3$-rd query, only Box $4$ is available. By putting Baggage $1$ into Box $4$, we can make the total value of baggage in boxes $9$, which is the maximum possible result.
题目信息
题号
abc195_d
时间限制
2 s
内存限制
1024 MB

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