思路：

我们浅推一波可以发现，每个格子看成一个点，向它能走到的点连边，然后就变成了最小覆盖问题
跑个匈牙利不成问题

$code$

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int dx[8] = {
     1, 1, 2, 2, -1, -1, -2, -2 };
const int dy[8] = {
     2, -2, 1, -1, 2, -2, 1, -1 };

int n, m, btot, wtot;
int a[25][25], color[25][25], link[5010], dian[25][25];
bool v[100010], ma[5010][5010];

bool find(int x) {
    
    for (int i = 1; i <= wtot; i++) {
    
        if (ma[x][i] == 1 && !v[i]) {
    
                v[i] = 1;
                int q = link[i];
                link[i] = x;
                if (q == 0 || find(q))
                    return 1;
                link[i] = q;
            }
    }
    return 0;
}

int main() {
    
    scanf("%d%d", &n, &m);
    int x, y, p = 0;
    while (cin >> x >> y) {
    
        a[x][y] = 1;
        p++;
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++) {
    
            if (a[i][j])
                continue;
            if ((i % 2 == 1 && j % 2 == 1) || (i % 2 == 0 && j % 2 == 0))
                color[i][j] = 1, dian[i][j] = ++ btot;
        	else dian[i][j] = ++ wtot;
		}
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (!a[i][j]) {
    
                for (int k = 0; k < 8; k++) {
    
                    int xx = i + dx[k], yy = j + dy[k];
                    if (xx < 1 || xx > n || yy < 1 || yy > m || a[xx][yy])
                        continue;
                    if (color[i][j] == 1)
                        ma[dian[i][j]][dian[xx][yy]] = 1;
                    else
                        ma[dian[xx][yy]][dian[i][j]] = 1;
                }
            }
    int ans = 0;
    for (int i = 1; i <= btot; i++) {
    
    	memset(v, 0, sizeof(v));
    	if(find(i)) ans ++;
    }
    printf("%d", btot + wtot - ans);
    return 0;
}