New Year and Domino
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 numbered 1 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 and c1i 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.
Examples
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.
题意：先给出一个矩阵，然后这些矩阵中有的有空，有的没空，每两个连续的空可以放一个多米诺骨牌，接下来给出一个查询q，对于每次查询题目给出一个子矩阵，求出每个子矩阵中可以放多少个多米诺骨牌，解题过程：首先先统计每个位置可以放几个多米诺骨牌（当前位置和右侧以及下侧）并将结果放进一个二维数组里，由于q很大，所以我们要用到二维前缀和，在格外开一个n*n的二维前缀数组，里面存放从（1,1）到当前位置假设为（x,y）所包含的矩阵的前缀和，对于每次给出的查询子矩阵，我们只需要用前缀和矩阵的相应位置进行操作即可
#include
#include
#include
#define sc(a) scanf("%d",&a)
#define scl(a) scanf("%lld",&a)
#define mem(a) memset(a,0,sizeof(a));
#define pr(a) printf("%d\n",a);
#define prl(a) printf("%lld\n",a);
using namespace std;
const int N = 505;
char ch[N][N];
int coun[N][N];
int pre[N][N];
int main()
{
memset(coun,0,sizeof(coun));
int h, w;
sc(h), sc(w);
//初始化二维数组
for (register int i = 0; i < N-1; i++) {
for (register int j = 0; j < N-1; j++) {
ch[i][j] = '#';
}
}
getchar();
//读取数据
for (int i = 1; i <= h; i++) {
for (int j = 1; j <= w; j++) {
scanf("%c", &ch[i][j]);
}
getchar();
}
//统计每个位置可以放几个骨牌
for (int i = 1; i <= h; i++) {
for (int j = 1; j <= w; j++) {
if (ch[i][j] == '.') {
if (ch[i][j + 1] == '.')
coun[i][j]++;
if (ch[i + 1][j] == '.')
coun[i][j]++;
}
}
}
// 制二维前缀数组
for (int i = 1; i <= h; i++) {
for (int j = 1; j <= w; j++) {
pre[i][j] = pre[i - 1][j] + pre[i][j - 1] -pre[i - 1][j - 1] + coun[i][j];
}
}
int q;
sc(q);
int x1, y1, x2, y2;
while (q--) {
sc(x1), sc(y1), sc(x2), sc(y2);
int ans = 0;
ans = pre[x2][y2] - pre[x2][y1 - 1] - pre[x1 - 1][y2] + pre[x1 - 1][y1 - 1];//利用前缀和性质
//我们还要讲下边界超出的点除去
for (int i = y1; i <= y2; i++) {
if (ch[x2][i] == '.'&&ch[x2 + 1][i] == '.') {
ans--;
}
}
//同理去除右边界
for (int i = x1; i <= x2; i++) {
if (ch[i][y2] == '.'&&ch[i][y2 + 1] == '.')
ans--;
}
pr(ans);
}
return 0;
}