#include <iostream>
#define int long long
using namespace std;
const short dir[8][2] = { {1 , 2} , {1 , -2} , {2 , 1} , {2 , -1} , {-1 , 2} , {-1 , -2} , {-2 , 1} , {-2 , -1} };
bool d[30][30];
int dp[30][30] , n , m , x , y;
signed main(){
    cin >> n >> m >> x >> y;
    d[x][y] = 1;
    for(int i = 0 ; i < 8 ; i++){
        int tx = x + dir[i][0],ty = y + dir[i][1];
        if(tx >=  0  &&  tx <= n  &&  ty >=  0  &&  ty <= m) 
			d[tx][ty] = 1;
    }
    dp[0][0] = 1;
    for(int i = 0 ; i <= n ; i++)
        for(int j = 0;j <= m;j++)
            if(!d[i][j]){
                if(i) 
					dp[i][j] +=  dp[i-1][j];
                if(j) 
					dp[i][j] +=  dp[i][j-1];
            }
    cout << dp[n][m] << endl;
    return 0;
}

#include<iostream>
using namespace std;
const int N=20+10;
int n,m,x,y;
bool v[N][N];
long long a[N][N];
int dx[]={0,-1,-2,-2,-1,1,2,2,1};
int dy[]={0,-2,-1,1,2,2,1,-1,-2};
void init(){
	for(int i=0;i<=8;i++){
		int xx=x+dx[i];
		int yy=y+dy[i];
		if(xx>=0&&xx<=n&&yy>=0&&yy<=n){
			v[xx][yy]=1;
		}
	}
	for(int i=0;i<=m;i++){
		if(v[0][i]==0){
			a[0][i]=1;
		}else{
			break;
		}
	}
	for(int i=0;i<=n;i++){
		if(v[i][0]==0){
			a[i][0]=1;
		}else{
			break;
		}
	}
}
int main(){
	cin>>n>>m>>x>>y;
	init();
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(v[i][j]==0){
				a[i][j]=a[i-1][j]+a[i][j-1];
			}
		}
	}
	cout<<a[n][m];
}

一道比较入门的dp题. 答案可能很大，所以记得开 long long.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int fx[9] = {0, -2, -1, 1, 2, 2, 1, -1, -2};
const int fy[9] = {0, 1, 2, 2, 1, -1, -2, -2, -1};
int bx, by, mx, my;
long long f[40][40];
bool s[40][40];
int main()
{
    scanf("%d%d%d%d",&bx,&by,&mx,&my);
    bx=bx+2; 
    by=by+2; 
    mx=mx+2;
    my=my+2;
    f[2][1]=1;
    s[mx][my]=1;
    for(int i=1;i<=8;i=i+1)s[mx+fx[i]][my+fy[i]]=1;
    for(int i=2;i<=bx;i=i+1)
    {
        for(int j=2;j<=by;j=j+1)
        {
            if(s[i][j]) continue;
            f[i][j]=f[i-1][j]+f[i][j-1];
        }
    }
    printf("%lld\n", f[bx][by]);
    return 0;
}