Good Bye 2015 C New Year and Domino(dp)

They say "years are like dominoes, tumbling one after the other". But would a year fit into a grid? I don't think so.

Limak is a little polar bear who loves to play. He has recently got a rectangular grid with h rows and w columns. Each cell is a square, either empty (denoted by '.') or forbidden (denoted by '#'). Rows are numbered 1 through h from top to bottom. Columns are numbered1 through w from left to right.

Also, Limak has a single domino. He wants to put it somewhere in a grid. A domino will occupy exactly two adjacent cells, located either in one row or in one column. Both adjacent cells must be empty and must be inside a grid.

Limak needs more fun and thus he is going to consider some queries. In each query he chooses some rectangle and wonders, how many way are there to put a single domino inside of the chosen rectangle?

Input

The first line of the input contains two integers h and w (1 ≤ h, w ≤ 500) – the number of rows and the number of columns, respectively.

The next h lines describe a grid. Each line contains a string of the length w. Each character is either '.' or '#' — denoting an empty or forbidden cell, respectively.

The next line contains a single integer q (1 ≤ q ≤ 100 000) — the number of queries.

Each of the next q lines contains four integers r1i, c1i, r2i, c2i (1 ≤ r1i ≤ r2i ≤ h, 1 ≤ c1i ≤ c2i ≤ w) — the i-th query. Numbers r1i andc1i denote the row and the column (respectively) of the upper left cell of the rectangle. Numbers r2i and c2i denote the row and the column (respectively) of the bottom right cell of the rectangle.

Output

Print q integers, i-th should be equal to the number of ways to put a single domino inside the i-th rectangle.

Sample test(s)

input
5 8
....#..#
.#......
##.#....
##..#.##
........
4
1 1 2 3
4 1 4 1
1 2 4 5
2 5 5 8

output
4
0
10
15

input
7 39
.......................................
.###..###..#..###.....###..###..#..###.
...#..#.#..#..#.........#..#.#..#..#...
.###..#.#..#..###.....###..#.#..#..###.
.#....#.#..#....#.....#....#.#..#..#.#.
.###..###..#..###.....###..###..#..###.
.......................................
6
1 1 3 20
2 10 6 30
2 10 7 30
2 2 7 7
1 7 7 7
1 8 7 8

output
53
89
120
23
0
2

Note

A red frame below corresponds to the first query of the first sample. A domino can be placed in 4 possible ways.

$\text{[math]}$

题目链接:点击打开链接

给出一个由'*' '.'构成的图, 进行q次询问, 每次询问给出两个点, 问这两个点间有多少'.'连在一块, 上下或者左右代表连在一块.

用两个dp数组保存结果, 1个记录左右的'.', 1个记录上下的'.', 每次询问通过两数组计算相邻的'.'就可以.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 505;
char maze[MAXN][MAXN];
int h, w, q, dp1[MAXN][MAXN], dp2[MAXN][MAXN];
int main(int argc, char const *argv[])
{
	scanf("%d%d", &h, &w);
	for(int i = 1; i <= h; ++i) {
		scanf("%s", maze[i] + 1);
		for(int j = 1; j <= w; ++j) {
			if(maze[i][j] == '.' && maze[i - 1][j] == '.') dp1[i][j]++;
			if(maze[i][j] == '.' && maze[i][j - 1] == '.') dp2[i][j]++;
			dp1[i][j] += dp1[i - 1][j] + dp1[i][j - 1] - dp1[i - 1][j - 1];
			dp2[i][j] += dp2[i - 1][j] + dp2[i][j - 1] - dp2[i - 1][j - 1];
		}
	}
	scanf("%d", &q);
	while(q--) {
		int x1, y1, x2, y2;
		scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
		printf("%d
", dp1[x1][y1 - 1] + dp1[x2][y2] - dp1[x1][y2] - dp1[x2][y1 - 1] 
			+ dp2[x1 - 1][y1] + dp2[x2][y2] - dp2[x2][y1] - dp2[x1 - 1][y2]);
	}
	return 0;
}