Good Bye 2015 C. New Year and Domino 二维前缀

C. New Year and Domino

 

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 numbered1 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 andc1i 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：  给你一个n*m的图，有 . , #  ，#号不可走

　　　　q个询问

　　　　每个询问对于一个矩形，问的就是 在这个矩形内 有多少条 路径为长度2 的路

题解：我们能求出每一个点 是否可以向下向右走的 即值为 2，1，0，

　　　　那么从 1，1到 x，y 这个矩形内可以用二维前缀数组求出来

　　　　每次询问 答案就是  简单容斥了，注意  边界及 只有 两个方向的情况

//meek
#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include<map>
#include<queue>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair

const int N=550+100;
const ll INF = 1ll<<61;
const int inf = 1000000007;
const int MOD= 1000007;


char mp[N][N];
int p[N][N],V[N][N];
int n,m;
int sss[4][2] ={1,0,0,1};
struct ss{
  int x,y;
};
int check(int x,int y) {
   if(x<=0||y<=0||x>n||y>m) return 1;
   return 0;
}
int v[N][N];
void bfs(int x,int y) {
    for(int i=1;i<=n;i++) {
        for(int j=1;j<=m;j++) {
                if(mp[i][j]!='#')
             for(int k=0;k<2;k++) {
                 int xx = i+sss[k][0];
                 int yy = j+sss[k][1];
                 if(check(xx,yy)) continue;
                 if(mp[xx][yy]=='#') continue;
                 p[i][j]++;
             }

        }
    }
}
int main() {
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) {
        getchar();
        for(int j=1;j<=m;j++) {
            scanf("%c",&mp[i][j]);
        }
    }
     bfs(1,1);


     for(int i=1;i<=n;i++) {
        for(int j=1;j<=m;j++) {
            V[i][j] = V[i][j]+V[i-1][j]+V[i][j-1]-V[i-1][j-1]+p[i][j];
        }
     }
    int q,x,y,x2,y2;
    scanf("%d",&q);
    for(int i=1;i<=q;i++) {
        scanf("%d%d%d%d",&x,&y,&x2,&y2);
        int  ans = V[x2][y2]-V[x2][y-1]-V[x-1][y2]+V[x-1][y-1];
        for(int j=x;j<x2;j++) {
            if(mp[j][y2]=='.'&&mp[j][y2+1]=='.') ans--;
        }
        ans-=p[x2][y2];
        for(int j=y;j<y2;j++) {
            if(mp[x2][j]=='.'&&mp[x2+1][j]=='.') ans--;
        }
        printf("%d\n",ans);
    }
}