C. New Year and Domino
题目连接：
http://www.codeforces.com/contest/611/problem/C
Description
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.
Sample Input
5 8
....#..#
.#......
.#....
..#.##
........
4
1 1 2 3
4 1 4 1
1 2 4 5
2 5 5 8
Sample Output
4
0
10
15
Hint
题意：
给你一个500 500的矩形区域，然后有一些位置是不能放的。有Q次询问，每次询问问你在(x1,y1,x2,y2)矩形区域中，能放入多少个1*2的矩形块。
题解
我们横着和竖着的都分开考虑。横着和竖着能放多少个呢？我们维护前缀和去解决就好了，如果上一位是空位，并且这一位置也是空位，就可以+1。n^2预处理，O(500)的单次询问。
代码
#include<bits/stdc++.h>
using namespace std;
int h,w;
char s[505][505];
int num[505][505];
int A[505][505];
int B[505][505];
int main()
{
cin>>h>>w;
for(int i=1;i<=h;i++)
scanf("%s",s[i]+1);
for(int i=1;i<=h;i++)
for(int j=1;j<=w;j++)
if(s[i][j]=='.')num[i][j]=1;
else num[i][j]=0;
for(int i=1;i<=h;i++)
{
for(int j=1;j<=w;j++)
{
if(num[i][j]&&num[i][j+1])A[i][j]=A[i][j-1]+1;
else A[i][j]=A[i][j-1];
}
}
for(int i=1;i<=w;i++)
{
for(int j=1;j<=h;j++)
{
if(num[j][i]&&num[j+1][i])B[j][i]=B[j-1][i]+1;
else B[j][i]=B[j-1][i];
}
}
int q;cin>>q;
while(q--)
{
int x1,y1,x2,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
int ans = 0;
for(int i=x1;i<=x2;i++)
ans+=A[i][y2-1]-A[i][y1-1];
for(int i=y1;i<=y2;i++)
ans+=B[x2-1][i]-B[x1-1][i];
printf("%d
",ans);
}
}