Codeforces 373D Counting Rectangles is Fun【Dp】

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本文介绍了一种算法，用于解决在一个N*M的矩阵中快速查询特定区间内全由0组成的子矩阵的数量问题。通过预处理的方式实现了高效查询。

There is an n × m rectangular grid, each cell of the grid contains a single integer: zero or one. Let's call the cell on the i-th row and the j-th column as (i, j).

Let's define a "rectangle" as four integers a, b, c, d (1 ≤ a ≤ c ≤ n; 1 ≤ b ≤ d ≤ m). Rectangle denotes a set of cells of the grid{(x, y) : a ≤ x ≤ c, b ≤ y ≤ d}. Let's define a "good rectangle" as a rectangle that includes only the cells with zeros.

You should answer the following q queries: calculate the number of good rectangles all of which cells are in the given rectangle.

Input

There are three integers in the first line: n, m and q (1 ≤ n, m ≤ 40, 1 ≤ q ≤ 3·105). Each of the next n lines contains m characters — the grid. Consider grid rows are numbered from top to bottom, and grid columns are numbered from left to right. Both columns and rows are numbered starting from 1.

Each of the next q lines contains a query — four integers that describe the current rectangle, a, b, c, d (1 ≤ a ≤ c ≤ n; 1 ≤ b ≤ d ≤ m).

Output

For each query output an answer — a single integer in a separate line.

Note

For the first example, there is a 5 × 5 rectangular grid, and the first, the second, and the third queries are represented in the following image.

$\text{[math]}$

For the first query, there are 10 good rectangles, five 1 × 1, two 2 × 1, two 1 × 2, and one 1 × 3.
For the second query, there is only one 1 × 1 good rectangle.
For the third query, there are 7 good rectangles, four 1 × 1, two 2 × 1, and one 3 × 1.

题目大意：

给出一个N*M的矩阵，有Q个查询，每个查询表示询问区间内有多少个全0子矩阵。

思路：

设定Dp【i】【j】【k】【l】表示区间（【a，b】【c，d】）内全0子矩阵的个数。

那么状态转移方程有：

Dp【i】【j】【k】【l】=Dp【i】【j】【k-1】【l】+Dp【i】【j】【k】【l-1】-Dp【i】【j】【k-1】【l-1】+以点（k，l）结尾能够组成的全0子矩阵的个数。

那么O（n^2*m^2）预处理然后O（1）查询答案即可。

Ac代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int row[55][55];
char a[55][55];
int dp[55][55][55][55];
int main()
{
    int n,m,q;
    while(~scanf("%d%d%d",&n,&m,&q))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%s",a[i]+1);
        }
        memset(dp,0,sizeof(dp));
        memset(row,0,sizeof(row));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(a[i][j]=='0')
                row[i][j]=row[i][j-1]+1;
                else row[i][j]=0;
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                for(int k=i;k<=n;k++)
                {
                    for(int l=j;l<=m;l++)
                    {
                        dp[i][j][k][l]=dp[i][j][k][l-1]+dp[i][j][k-1][l]-dp[i][j][k-1][l-1];
                        int r=l-j+1;
                        for(int z=k;z>=i;z--)
                        {
                            r=min(r,row[z][l]);
                            dp[i][j][k][l]+=r;
                        }
                    }
                }
            }
        }
        while(q--)
        {
            int a,b,c,d;
            scanf("%d%d%d%d",&a,&b,&c,&d);
            printf("%d\n",dp[a][b][c][d]);
        }
    }
}