Codeforces Good Bye 2015 C. New Year and Domino (预处理)

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 andw columns. Each cell is a square, either empty (denoted by '.') or forbidden (denoted by '#'). Rows are numbered1 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 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 and c1_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
....#..#
.#......
##.#....
##..#.##
........
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：给你一个地图，有的地方不能放东西，有的地方能放东西，然后给你一些区域的对角两端坐标，求能放多少个

长度为2的东西

思路：预处理行和列从头开始到当前位置能放的木棍数，如果上一位是空位，并且这一位置也是空位，就可以+1，然后按行按列查找相加

总结：写的时候有点糊涂，改了好多次才改好，对上样例交了，没想到A了

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 60001
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int num[550][550];
char map[550][550];
int a[550][550];
int b[550][550];
int main()
{
	int n,m,i,j,t,x1,x2,y1,y2;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++)
		scanf("%s",map[i]+1);
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				if(map[i][j]=='.')
				num[i][j]=0;
				else
				num[i][j]=1;
			}
		}
		mem(a);mem(b);
		for(i=1;i<=n;i++)//按行处理到达每个位置能放的木棍数
		{
			for(j=2;j<=m;j++)
			{
				if(num[i][j-1]==0&&num[i][j]==0)
				a[i][j]=a[i][j-1]+1;
				else
				a[i][j]=a[i][j-1];
			}
		}
		for(i=1;i<=m;i++)//按列处理到达每个位置能放的木棍数
		{
			for(j=2;j<=n;j++)
			{
				if(num[j-1][i]==0&&num[j][i]==0)
				b[j][i]=b[j-1][i]+1;
				else
				b[j][i]=b[j-1][i];
			}
		}
		scanf("%d",&t);
		while(t--)
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			ll ans=0;
//			for(i=1;i<=m;i++)
//			printf("%d ",a[1][i]);
			for(i=x1;i<=x2;i++)
			{
				ans+=a[i][y2]-a[i][y1];
				//printf("%d %d\n",a[i][y2],a[i][y1]);
			}
			//printf("%I64d\n",ans);
			for(i=y1;i<=y2;i++)
			{
				ans+=b[x2][i]-b[x1][i];
			}
			printf("%I64d\n",ans);
		}
	}
	return 0;
}