计算机学院大学生程序设计竞赛（2015’12）01 Matrix

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im0qianqian 2022-08-11 00:51:57 ©著作权

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01 Matrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 564 Accepted Submission(s): 121

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

2 2 2 01 00 1 2 3 3 010 111 111 1 2 2

Sample Output

1 0 7 2 2

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#include <iostream>
#include<stdio.h>
#include <cstring>
using namespace std;
int a[1001][1001] ,b[1000];
char s[1001][1001];
int main()
{
    int t,n,m;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        int ma=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0; i<n; i++)
            scanf("%s",s[i]);
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                a[i+1][j+1]=s[i][j]-'0';
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                if(a[i][j]==1)
                {
                    int mi=a[i-1][j-1];
                    if(mi>a[i-1][j])mi=a[i-1][j];
                    if(mi>a[i][j-1])mi=a[i][j-1];
                    mi++;
                    a[i][j]=mi;
                    b[mi]++;
                    ma=max(mi,ma);
                }
            }
        for(int i=ma; i>1; i--)
        {
            b[i-1]+=b[i];
        }
        int x;
        for(int i=0; i<m; i++)
        {
            cin>>x;
            cout<<b[x]<<endl;
        }
    }
    return 0;
}