Educational Codeforces Round 89 (Rated for Div. 2) C. Palindromic Paths

C. Palindromic Paths
time limit per test1.5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a matrix with n rows (numbered from 1 to n) and m columns (numbered from 1 to m). A number ai,j is written in the cell belonging to the i-th row and the j-th column, each number is either 0 or 1.

A chip is initially in the cell (1,1), and it will be moved to the cell (n,m). During each move, it either moves to the next cell in the current row, or in the current column (if the current cell is (x,y), then after the move it can be either (x+1,y) or (x,y+1)). The chip cannot leave the matrix.

Consider each path of the chip from (1,1) to (n,m). A path is called palindromic if the number in the first cell is equal to the number in the last cell, the number in the second cell is equal to the number in the second-to-last cell, and so on.

Your goal is to change the values in the minimum number of cells so that every path is palindromic.

Input
The first line contains one integer t (1≤t≤200) — the number of test cases.

The first line of each test case contains two integers n and m (2≤n,m≤30) — the dimensions of the matrix.

Then n lines follow, the i-th line contains m integers ai,1, ai,2, …, ai,m (0≤ai,j≤1).

Output
For each test case, print one integer — the minimum number of cells you have to change so that every path in the matrix is palindromic.

Example
inputCopy
4
2 2
1 1
0 1
2 3
1 1 0
1 0 0
3 7
1 0 1 1 1 1 1
0 0 0 0 0 0 0
1 1 1 1 1 0 1
3 5
1 0 1 0 0
1 1 1 1 0
0 0 1 0 0
outputCopy
0
3
4
4
Note
The resulting matrices in the first three test cases:

			   	 (1011)
			    (000000)
		（101011111101111110101）

int t;
int ans;
int m,n;

int main() {
    cin>>t;
    while(t--) {
        int a[35][35];
        cin>>n>>m;
        ans=0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
                cin>>a[i][j];
        for(int length=0; length<=(n+m-2); length++) {
            if(2*length==n+m-2)
                continue;
            int ones=0,zeros=0;
            for(int i=1; i<=n; i++) {
                if(length-i+2>=1 and length-i+2<=m) {
                    if(a[i][length-i+2]==1)
                        ones++;
                    else
                        zeros++;

                }

            }
            for(int i=1; i<=n; i++) {
                if(n+m-length-i>=1 && n+m-length-i<=m) {
                    if(a[i][n+m-length-i]==1)
                        ones++;
                    else
                        zeros++;
                }
            }
            ans=ans+min(zeros,ones);
        }
        cout<<ans/2<<endl;
    }

}