Codeforces 611C

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h,w表示行和列的数目，接下来h行描述一个网格，每一行有一个长度为w的字符串，每个字符要么是'.',要么是'#'，分别表示空的和禁止的。

　　输入一个q，接下来得q行有4个数，表示一个小网格的左上角的横纵坐标和右下角的横纵坐标。

　　输出每个小方格内有几个2个连在一起的空格。

思路：网格存在gg[MAXN][MAXN]中，用两个二位数组row[MAXN][MAXN]和column[MAXN][MAXN];row[i][j]表示第i行从开始到j这个位置有几个2个连在一起的空格，column[i][j]表示第j列从开始到i这个位置有几个2个连在一起的空格。

row举例来说，如果gg[i][j]这个位置是'.'并且它前面gg[i][j-1]这个位置也是'.'的话row[i][j]=row[i][j-1]+1；否则的话，row[i][j]=row[i][j-1]。

要注意越界的问题。

if(gg[i][j]=='.'&&j>1&&gg[i][j-1]=='.') row[i][j]=row[i][j-1]+1;
else row[i][j]=row[i][j-1];

if(gg[i][j]=='.'&&i>1&&gg[i-1][j]=='.') column[i][j]=column[i-1][j]+1;
else column[i][j]=column[i-1][j];

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char gg[510][510];
int row[510][510],column[510][510];
int x1[100010],y1[100010],x2[100010],y2[100010];
int main()
{
    int h,w,q;
    scanf("%d %d",&h,&w);
    getchar();
    memset(row,0,sizeof(row));
    memset(column,0,sizeof(column));
    int i,j;
    for(i=1; i<=h; i++)
    {
        for(j=1; j<=w; j++)
        {
            scanf("%c",&gg[i][j]);
            if(gg[i][j]=='.'&&j>1&&gg[i][j-1]=='.') row[i][j]=row[i][j-1]+1;
            else row[i][j]=row[i][j-1];
            if(gg[i][j]=='.'&&i>1&&gg[i-1][j]=='.') column[i][j]=column[i-1][j]+1;
            else column[i][j]=column[i-1][j];
        }
        getchar();
    }
    /*
    for(i=1; i<=h; i++)
    {
        for(j=1; j<=w; j++)
            cout<<row[i][j]<<" ";
        cout<<endl;
    }
    cout<<endl<<endl;
    for(i=1; i<=h; i++)
    {
        for(j=1; j<=w; j++)
            cout<<column[i][j]<<" ";
        cout<<endl;
    }
    */
    scanf("%d",&q);
    for(i=0; i<q; i++)
        scanf("%d%d%d%d",&x1[i],&y1[i],&x2[i],&y2[i]);
    __int64 ans;
    for(i=0; i<q; i++)
    {
        ans=0;
        for(j=x1[i]; j<=x2[i]; j++)
        {
            //cout<<row[j][y2[i]]<<" "<<row[j][y1[i]]<<endl;
            ans+=row[j][y2[i]]-row[j][y1[i]];
        }
        for(j=y1[i]; j<=y2[i]; j++)
        {
            //cout<<column[x2[i]][j]<<" "<<column[x1[i]][j]<<endl;
            ans+=column[x2[i]][j]-column[x1[i]][j];
        }
        cout<<ans<<endl;
    }

    return 0;
}