611C. New Year and Domino【二维前缀和】【容斥】

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 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....#..#.#......##.#....##..#.##........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.

这题主要是二维数组的和的预处理。

先扫一遍记录每个格点的种数，然后累加presum[i][j]+=presum[i-1][j]+presum[i][j-1]-presum[i-1][j-1];

这样presum保存的就是点(i,j)到(1,1)的矩形的和了，

计算子矩形的时候用到了容斥，cnt=presum[r2][c2]-presum[r1-1][c2]-presum[r2][c1-1]+presum[r1-1][c1-1];

最后扫一遍边缘把例外情况去掉就行了。

时间复杂度：h*w+q

#include <iostream>#include <cstring>#include <iomanip>using namespace std;char a[500+10][500+10];int presum[500+10][500+10]={0};bool f_right[500+10][500+10]={0};bool f_down[500+10][500+10]={0};int main(void){cin.tie(0);ios::sync_with_stdio(0);int h,w;cin>>h>>w;for(int i=1 ; i<=h ; ++i)for(int j=1 ; j<=w ; ++j)cin>>a[i][j];for(int i=1 ; i<=h ; ++i){for(int j=1 ; j<=w ; ++j){if(a[i][j]=='.' && a[i][j+1]=='.'){f_right[i][j]=1;presum[i][j]++;}if(a[i][j]=='.' && a[i+1][j]=='.'){f_down[i][j]=1;presum[i][j]++;}presum[i][j]+=presum[i-1][j]+presum[i][j-1]-presum[i-1][j-1];}}int q;cin>>q;while(q--){int r1,c1,r2,c2;cin>>r1>>c1>>r2>>c2;int cnt=presum[r2][c2]-presum[r1-1][c2]-presum[r2][c1-1]+presum[r1-1][c1-1];for(int i=r1 ; i<=r2 ; ++i)if(f_right[i][c2])cnt--;for(int j=c1 ; j<=c2 ; ++j)if(f_down[r2][j])cnt--;cout<<cnt<<"\n";}return 0;}