【Goodbye2015】Codeforces 611C New Year and Domino【思维】

最新推荐文章于 2020-06-30 15:26:55 发布

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最新推荐文章于 2020-06-30 15:26:55 发布

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分类专栏：思维文章标签： Codeforces 611C Goodbye2015

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本篇介绍了一个关于在指定矩形区域内计数多米诺骨牌不同放置方式的问题，通过预处理二维前缀和来高效解答多次查询。

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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 r1_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. Numbers r2_i and c2_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 the i-th rectangle.

Examples

Input

5 8
....#..#
.#......
##.#....
##..#.##
........
4
1 1 2 3
4 1 4 1
1 2 4 5
2 5 5 8

Output

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

Note

A red frame below corresponds to the first query of the first sample. A domino can be placed in 4 possible ways.

$\text{[math]}$

题目大意：

给你一个N*M的图，其中有两个相邻的“.”，就可以放置这个多股诺骨牌。

一共Q个询问，每个询问给出一个子矩阵，问你这个子矩阵中有多少种摆放的方式。

思路：

1、直接维护一个区间的和可能相对较为麻烦一些，我们不妨将问题分割成两部分：
①横着放。

②竖着放。

2、那么对于两个子问题，我们分别维护两个二维前缀和：

sum【i】【j】表示第i行，从位子1到位子j横着放置的方案数。

sum2【i】【j】表示第j列，从位子1到位子i竖着放置的方案数。

那么对于查询，我们枚举每一行进行加和，再枚举每一列进行加和。

时间复杂度O((n+m)*q);

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[505][505];
int sum2[505][505];
int sum[505][505];
int n,m;
void init()
{
    memset(sum,0,sizeof(sum));
    memset(sum2,0,sizeof(sum2));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(j==1)sum[i][j]=0;
            else
            {
                int tmp;
                if(a[i][j]=='.'&&a[i][j-1]=='.')tmp=1;
                else tmp=0;
                sum[i][j]=sum[i][j-1]+tmp;
            }
        }
    }
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(j==1)sum2[j][i]=0;
            else
            {
                int tmp;
                if(a[j][i]=='.'&&a[j-1][i]=='.')tmp=1;
                else tmp=0;
                sum2[j][i]=sum2[j-1][i]+tmp;
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)scanf("%s",a[i]+1);
        init();
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            int output=0;
            for(int i=x1;i<=x2;i++)output+=sum[i][y2]-sum[i][y1];
            for(int i=y1;i<=y2;i++)output+=sum2[x2][i]-sum2[x1][i];
            printf("%d\n",output);
        }
    }
}