Codeforces 611C New Year and Domino（dp）

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.

分析：

d[i][j]数组保存从(1,1）——(i,j)的可行情况数,

#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=500+9;
#define inf 0x3f3f3f3f
int h,w,d[maxn][maxn];
char a[maxn][maxn];
int main()
{
   // freopen("f.txt","r",stdin);
    scanf("%d%d",&h,&w);
    for(int i=1;i<=h;i++)scanf("%s",a[i]+1);
    memset(d,0,sizeof(d));

    for(int i=2;i<=h;i++)
        if(a[i][1]=='.'&&a[i-1][1]=='.')d[i][1]=d[i-1][1]+1;
        else d[i][1]=d[i-1][1];
    for(int i=2;i<=w;i++)
        if(a[1][i]=='.'&&a[1][i-1]=='.')d[1][i]=d[1][i-1]+1;
        else d[1][i]=d[1][i-1];
    int temp;
    for(int i=2;i<=h;i++){
        for(int j=2;j<=w;j++){
            temp=0;
            if(a[i][j]=='#')temp=d[i-1][j]+d[i][j-1]-d[i-1][j-1];
            else {
                if(a[i][j-1]=='.'&&a[i-1][j]=='.')temp=d[i-1][j]+d[i][j-1]-d[i-1][j-1]+2;
                else if(a[i][j-1]=='#'&&a[i-1][j]=='#')temp=d[i-1][j]+d[i][j-1]-d[i-1][j-1];
                else temp=d[i-1][j]+d[i][j-1]-d[i-1][j-1]+1;
            }
            d[i][j]=temp;
        //d[i][j]=max(temp,d[i][j]);
        }
    }
    int n,r1,r2,s1,s2;
    scanf("%d",&n);
    while(n--){
        scanf("%d%d%d%d",&r1,&s1,&r2,&s2);
        int sum=0;
        for(int i=s1;i