C. New Year and Domino
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

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.

题意：输入一个矩阵  '.'代表空地  q个询问 每个询问 两组坐标 求其构成的长方形   中如上图占据两个空格  有多少种方法
题解：C题耗了好久  重点在预处理 遍历一遍矩阵 只 向左 向右两个方向 寻找满足的情况 并累加
同时更新 anss[i][j]=ans-anss[i-1][d]+anss[i-1][j]; anss[i][j] 代表（1，1）与（i，j）组成的矩阵中有多少种方法
输出询问的结果re=anss[r2][c2]-anss[r1-1][c2]-anss[r2][c1-1 ]+anss[r1-1][c1-1]-flag;
重点说一下 flag是什么？ 为询问的矩阵的左边向左与上边向上 能构成一种方法的总数
SAY goodbye 2015
24小时就睡了2小时 和基地友出去玩
还是想花时间 码个日志！！！


#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#define LL __int64
using namespace std;
int h,w;
int q;
char mp[505][505];
int dis[2][2]={{0,-1},{-1,0}};
int gg[505][505];
int anss[505][505];
void  fun(int a,int b,int c,int d)
{
    int ans=0;
    for(int i=a;i<=c;i++)
        for(int j=b;j<=d;j++)
    {
        if(mp[i][j]=='.')
        {
            for(int k=0;k<2;k++)
            {
                int xx=i+dis[k][0];
                int yy=j+dis[k][1];
                if(xx>=a&&xx<=c&&yy>=b&&yy<=d&&mp[xx][yy]=='.')
                    ans++;
            }
        }
         anss[i][j]=ans-anss[i-1][d]+anss[i-1][j];
    }
}
int main()
{
    scanf("%d%d",&h,&w);

    for(int i=1; i<=h; i++)
    {
        getchar();
        for(int j=1; j<=w; j++)
            scanf("%c",&mp[i][j]);
    }
    fun(1,1,h,w);
    scanf("%d",&q);
    for(int i=1;i<=q;i++)
    {   int r1,c1,r2,c2;
        scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
        int flag=0;
        for(int k=c1;k<=c2;k++)
        {
            if(mp[r1][k]=='.')
            {
                if(mp[r1-1][k]=='.')
                    flag++;
            }
        }
        for(int k=r1;k<=r2;k++)
        {
            if(mp[k][c1]=='.')
            {
                if(mp[k][c1-1]=='.')
                    flag++;
            }
        }
        //cout<<flag<<endl;
        printf("%d\n",anss[r2][c2]-anss[r1-1][c2]-anss[r2][c1-1 ]+anss[r1-1][c1-1]-flag);
    }
    return 0;
}