P1002 [NOIP2002 普及组] 过河卒

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。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int n, m, a, b;
int dx[9] = {0, -2, -1, 1, 2, 2, 1, -1, -2};
int dy[9] = {0, -1, -2, -2, -1, 1, 2, 2, 1};
int mp[30][30];
ll f[30][30];//路径很多，要开longlong 
bool check(int x, int y) {
    if ((x >= 1) && (x <= n) && (y >= 1) && (y <= m)) return 1;
    else return 0;
}
int main() {
    scanf("%d %d %d %d", &n, &m, &a, &b);
    n ++, m ++, a ++, b ++;//下标从1开始 
    mp[a][b] = 1;
    for (int i = 1; i <= 8; i ++) {
        int nx = a + dx[i], ny = b + dy[i];
        if (check(nx, ny)) mp[nx][ny] = 1;
    }
    f[1][1] = 1;
    for (int i = 1; i <= n; i ++) {
        for (int j = 1; j <= m; j ++){
            if (i == 1 && j == 1) continue;
            if (mp[i][j]) continue;
            f[i][j] = f[i][j - 1] + f[i - 1][j];
        }
    }
    printf("%lld\n", f[n][m]);
    return 0;        
}