计算机学院大学生程序设计竞赛（2015’12） 1006 01 Matrix(dp)

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本文介绍了一种高效算法，用于计算给定01矩阵中所有大小为k*k的全1子矩阵的数量。通过动态规划方法，预先计算每个位置上可能存在的最大全1子矩阵尺寸，并使用哈希表进行快速查询，从而实现O(n^2)时间复杂度内的求解。

01 Matrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 605 Accepted Submission(s): 127

Problem Description

It's really a simple problem.
Given a "01" matrix with size by n*n (the matrix size is n*n and only contain "0" or "1" in each grid), please count the number of "1" matrix with size by k*k (the matrix size is k*k and only contain "1" in each grid).

Input

There is an integer T (0 < T <=50) in the first line, indicating the case number.
Each test case begins with two numbers n and m (0<n, m<=1000), specifying the size of matrix and the query number.
Then n lines follow and each line contains n chars ("0" or "1").
Then m lines follow, each lines contains a number k (0<k<=n).

Output

For each query, output the number of "1" matrix with size by k*k.

Sample Input

Sample Output

1
0
7
2
2
首先要感谢金巨的指导
题解：
dp[i][j]代表着点（i，j）右下方的最长可行矩阵的长度
L[i][j]代表着点（i，j）右方的最长可行矩阵的长度
R[i][j]代表着点（i，j）下方的最长可行矩阵的长度

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<set>
#include<stack>
#include<map>
#include<algorithm>
#include<list>
#include<cmath>
#define LL long long
#define INF 0x3f3f3f3f
#define eps 1e-8
using namespace std;

int dp[1010][1010],L[1010][1010],R[1010][1010],Hash[1010];
char str[1010][1010];

int main()
{
    int T,n,m,k;
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T-- && scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n+1; i++)
            for(int j=1; j<=n+1; j++)
                dp[i][j] = L[i][j] = R[i][j] =  Hash[i] = 0;
        for(int i=1; i<=n; i++)
            scanf("%s",str[i]+1);
        for(int i=n; i>=1; i--)
        {
            for(int j=n; j>=1; j--)
            {
                if(str[i][j] == '0')
                    dp[i][j] = L[i][j] = R[i][j] = 0;
                else
                {
                    L[i][j] = L[i+1][j] + 1;
                    R[i][j] = R[i][j+1] + 1;
                    dp[i][j] = min(min(L[i][j],R[i][j]),dp[i+1][j+1]+1);
                    Hash[dp[i][j]]++;
                }
            }
        }
        for(int i=n; i>=1; i--)
            Hash[i] += Hash[i+1];
        while(m--)
        {
            scanf("%d",&k);
            printf("%d\n",Hash[k]);
        }
    }

    return 0;
}