P1002 

标记马可以到达的地方，因为卒是能向下或向右走，设f[i][j]表示到达(i,j)的路径数，显然有：f[i][j] = f[i - 1][j] + f[i][j - 1]。

DP转移即可。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 int n, m, a, b;
 5 int dx[9] = {0, -2, -1, 1, 2, 2, 1, -1, -2};
 6 int dy[9] = {0, -1, -2, -2, -1, 1, 2, 2, 1};
 7 int mp[30][30];
 8 ll f[30][30];//路径很多，要开longlong 
 9 bool check(int x, int y) {
10     if ((x >= 1) && (x <= n) && (y >= 1) && (y <= m)) return 1;
11     else return 0;
12 }
13 int main() {
14     scanf("%d %d %d %d", &n, &m, &a, &b);
15     n ++, m ++, a ++, b ++;//下标从1开始 
16     mp[a][b] = 1;
17     for (int i = 1; i <= 8; i ++) {
18         int nx = a + dx[i], ny = b + dy[i];
19         if (check(nx, ny)) mp[nx][ny] = 1;
20     }
21     f[1][1] = 1;
22     for (int i = 1; i <= n; i ++) {
23         for (int j = 1; j <= m; j ++){
24             if (i == 1 && j == 1) continue;
25             if (mp[i][j]) continue;
26             f[i][j] = f[i][j - 1] + f[i - 1][j];
27         }
28     }
29     printf("%lld\n", f[n][m]);
30     return 0;        
31 }