시간 제한 | 메모리 제한 | 제출 | 정답 | 맞힌 사람 | 정답 비율 |
---|---|---|---|---|---|
3 초 | 256 MB | 6 | 6 | 6 | 100.000% |
You are given an $H \times W$ grid. The square at the top-left corner is indexed $(0, 0)$, and the square at the bottom-right corner is indexed by $(H-1, W-1)$.
$N$ squares $(x_1, y_1), (x_2, y_2), \ldots, (x_N, y_N)$ are painted black, and all other squares are painted white.
Let the shortest distance between white squares $A$ and $B$ be the minimum number of moves required to reach $B$ from $A$ visiting only white squares, where one can travel to an adjacent square sharing a side (up, down, left or right) in one move.
Since there are $H \times W - N$ white squares in total, there are $C_{H \times W - N}^2$ ways to choose two of the white squares. For each of these $C_{H \times W - N}^2$ ways, find the shortest distance between the chosen squares, then find the sum of all those distances, modulo $1\,000\,000\,007=10^9+7$.
Input is given in the following format:
$H$ $W$
$N$
$x_1$ $y_1$
$x_2$ $y_2$
$\ldots$
$x_N$ $y_N$
Print the sum of the shortest distances modulo $10^9+7$.
$1 \leq H, W \leq 10^6$, $1 \leq N \leq 30$, $0 \leq x_i \leq H-1$, $0 \leq y_i \leq W-1$. If $i \neq j$, then either $x_i \neq x_j$ or $y_i \neq y_j$. It is guaranteed that there is at least one white square. For every pair of white squares $A$ and $B$, it is possible to reach $B$ from $A$ visiting only white squares.
2 3 1 1 1
20
2 3 1 1 2
16
3 3 1 1 1
64
4 4 4 0 1 1 1 2 1 2 2
268
1000000 1000000 1 0 0
333211937
In Sample 1, we have the next grid ('.
' denotes white square, '!
' --- black square):
... .!.
We assign alphabet to white squares, like below.
ABC D!E
So we get (here $dist(A,B)$ is the shortest distance between $A$ and $B$):
$dist(A, B) = 1$, $dist(A, C) = 2$, $dist(A, D) = 1$, $dist(A, E) = 3$, $dist(B, C) = 1$, $dist(B, D) = 2$, $dist(B, E) = 2$, $dist(C, D) = 3$, $dist(C, E) = 1$, $dist(D, E) = 4$, and sum of those is $20$.
In Sample 2, we assign alphabet to white squares, like below.
ABC DE!
So we get:
$dist(A, B) = 1$, $dist(A, C) = 2$, $dist(A, D) = 1$, $dist(A, E) = 2$, $dist(B, C) = 1$, $dist(B, D) = 2$, $dist(B, E) = 1$, $dist(C, D) = 3$, $dist(C, E) = 2$, $dist(D, E) = 1$, and sum of those is $16$.