P1387 largest square

Topic description

Find the largest square that does not contain 0 in an n*m matrix containing only 0 and 1, and output the side length.

Input and output format

Input format:

The first line of the input file contains two integers n, m (1<=n, m<=100), the next n lines, each line contains m numbers, separated by spaces, 0 or 1.

Output format:

an integer, the side length of the largest square

Input and output example

Input Example #1: Copy

4 4
0 1 1 1
1 1 1 0
0 1 1 0
1 1 0 1

Output Sample #1: Copy

2

 

Solution：

　　This question is too watery ($Brave-Cattle$ giants ask me how to do this? ), direct two-dimensional prefixes and enumeration violence are $OK$.

　　$s[i][j]$ represents the sum of the numbers contained in the rectangle with $(1,1), (i,j)$ as the diagonal vertices, when the square with side length $a$ only contains $1$ , then the sum of the numbers in the square is $a^2$, so enumerate the side lengths from large to small, then enumerate the positions of the points, and judge $s[i][j]-s[i][ja ]-s[ia][j]+s[ia][ja]$ is equal to $a^2$. The time complexity $O(n^3)$ is completely passable.

Code:

 

 1 #include<bits/stdc++.h>
 2 #define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
 3 using namespace std;
 4 const int N=105;
 5 int n,m,s[N][N],tot,x;
 6 int main(){
 7     ios::sync_with_stdio(0);
 8     cin>>n>>m;
 9     For(i,1,n) {
10         tot=0;
11         For(j,1,m)cin>>x,tot+=x,s[i][j]+=s[i-1][j]+tot;
12     }
13     x=n>m?m:n;
14     while(x){
15         For(i,x,n) For(j,x,m)
16             if(s[i][j]-s[i][j-x]-s[i-x][j]+s[i-x][j-x]==x*x){cout<<x;return 0;}
17         x--;
18     }
19 }