Good Bye 2015 C

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 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_i 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 201524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
",anss[r2][c2]-anss[r1-1][c2]-anss[r2][c1-1 ]+anss[r1-1][c1-1]-flag);
    }
    return 0;
}