New Year and Domino：二维前缀和

题目描述：

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 r1_i, c1_i, r2_i, c2_i (1 ≤ r1_i ≤ r2_i ≤ h, 1 ≤ c1_i ≤ c2_i ≤ w) — the i-th query. Numbers r1_i and c1_i denote the row and the column (respectively) of the upper left cell of the rectangle. Numbers r2_i and c2_idenote 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.

题目大意：n行m列，q次询问，然后问询问区间内 左右两连点 和上下两连点有多少个。

a图

最大的矩形前缀和就等于蓝的矩阵加上绿的矩阵，再减去重叠面积，最后加上小方块，即

sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + a[i][j]

引自

https://www.cnblogs.com/mrclr/p/8423136.html

b图

求红色方块面积 就是s1-s2-s3+s4

#include<iostream>
using namespace std;
char a[510][510];
int x[510][510],y[510][510],m,n,q,x1,x2,y1,y2,ans;
int main()
 {
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    cin>>a[i][j];

    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
        if(a[i][j]=='.')
        {
            if(a[i-1][j]=='.')x[i][j]++;
            if(a[i][j-1]=='.')y[i][j]++;
        }
        x[i][j]+=x[i-1][j]+x[i][j-1]-x[i-1][j-1];//前缀和的重要公式   看a图
        y[i][j]+=y[i-1][j]+y[i][j-1]-y[i-1][j-1];//前缀和的重要公式   看a图
    }
    cin>>q;
    while(q--)
    {
        ans=0;
        cin >> x1 >> y1 >> x2 >> y2;
        ans+= x[x2][y2] - x[x1][y2] - x[x2][y1 - 1] + x[x1][y1-1];//看b图
        ans+= y[x2][y2] - y[x2][y1] - y[x1 - 1][y2] + y[x1-1][y1];//看b图
        cout << ans << endl;
    }
    return 0;
}