codeforces-611C-New Year and Domino

611C-New Year and Domino

                    time limit per test3 seconds  memory limit per test256 megabytes

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 r1i, c1i, r2i, c2i (1 ≤ r1i ≤ r2i ≤ h, 1 ≤ c1i ≤ c2i ≤ w) — the i-th query. Numbers r1i and c1i 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.

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.

题目链接：cf-611C

题目大意：这道题目看note就比较好理解

题目思路：分别以行列dp，注意减去重复的区域

类似题目：nyoj-687（稍难）

以下是代码：

#include<iostream>
#include<string>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define ll long long
char a[505][505];
int dp[505][505];
int r[505][505], c[505][505];
int n, m;
int main(){
 cin >> n >> m;
 for( int i=1 ; i<=n ; i++ ){
 for (int j = 1; j <= m; j++)
 {

 cin >> a[i][j];
 }
 }
 for( int i=1 ; i<=n ; i++ ){

 for( int j=1 ; j<=m ; j++ ){
 if (a[i][j] == '.' && a[i - 1][j] == '.') r[i][j]++;
 if (a[i][j - 1] =='.' && a[i][j] == '.') c[i][j]++;
 r[i][j] += r[i][j - 1] + r[i - 1][j] - r[i - 1][j - 1];
 c[i][j] += c[i][j - 1] + c[i - 1][j]-c[i - 1][j - 1];
 }
 }
 int q;
 cin >> q;
 while(q--)
 {
 int x1,y1,x2,y2;
 cin >> x1 >> y1 >> x2 >> y2;
 int ans = r[x2][y2] - r[x1][y2] - r[x2][y1 - 1] + r[x1][y1 - 1] + c[x2][y2] - c[x2][y1] - c[x1 - 1][y2] + c[x1-1][y1];
 cout << ans << endl;
 }
 return 0;
}