Good Bye 2015 C. New Year and Domino dp+容斥原理

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 withh 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 andw (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 lengthw. 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 integersr1_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 andc1_i denote the row and the column (respectively) of the upper left cell of the rectangle. Numbersr2_i andc2_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 thei-th rectangle.

Sample test(s)

Input

5 8....#..#.#......##.#....##..#.##........41 1 2 34 1 4 11 2 4 52 5 5 8

Output

401015

Input

7 39........................................###..###..#..###.....###..###..#..###....#..#.#..#..#.........#..#.#..#..#....###..#.#..#..###.....###..#.#..#..###..#....#.#..#....#.....#....#.#..#..#.#..###..###..#..###.....###..###..#..###........................................61 1 3 202 10 6 302 10 7 302 2 7 71 7 7 71 8 7 8

Output

53891202302

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：给出一个图，q次询问，每次询问这个图的某一个矩形区域装下一个长为1宽为2的矩形的方法有多少种。

思路：dp+容斥原理

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;char Map[505][505];int dp[505][505];int h, w, q, r1, c1, r2, c2;int main(){    int i, j, res;    scanf("%d %d", &h, &w);    for(int i = 0;i < h;i++)        scanf("%s", Map[i]);    for(i = 0;i < h;i++)        if(i - 1 >= 0&&Map[i-1][0] == '.'&&Map[i][0] == '.') dp[i][0] = dp[i-1][0] + 1;        else dp[i][0] = dp[i-1][0];    for(i = 0;i < w;i++)        if(i - 1 >= 0&&Map[0][i-1] == '.'&&Map[0][i] == '.') dp[0][i] = dp[0][i-1] + 1;        else dp[0][i] = dp[0][i-1];    for(i = 1;i < h;i++)    {        for(j = 1;j < w;j++)        {            dp[i][j] += dp[i - 1][j];            dp[i][j] += dp[i][j - 1];            if(Map[i][j] == '.'){                if(Map[i-1][j] == '.') dp[i][j]++;                if(Map[i][j-1] == '.') dp[i][j]++;            }            dp[i][j] -= dp[i-1][j-1];        }    }    scanf("%d", &q);    while(q--){        scanf("%d %d %d %d", &r1, &c1, &r2, &c2);        //printf("%d %d", r2-r1-1, c2-c1-1);        res = dp[r2-1][c2-1] - dp[r1-1][c2-1] - dp[r2-1][c1-1] + dp[r1-1][c1-1];        for(i = r1-1;i <= r2-1;i++) if(i-1>=r1-1&&Map[i-1][c1-1]=='.'&&Map[i][c1-1]=='.') res++;        for(i = c1-1;i <= c2-1;i++) if(i-1>=c1-1&&Map[r1-1][i-1]=='.'&&Map[r1-1][i]=='.') res++;        printf("%d\n", res);    }}