51 lines
1.6 KiB
C++
51 lines
1.6 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 22;
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LL f[N][N]; //递推小心爆INT，不开LL见祖宗
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int ctrl[N][N]; //定义马的控制范围,一般采用int数组进行状态标识，不采用bool,因为1,0比true和false写的快
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int n, m, x, y; //B点(目标点)的坐标，马的坐标
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//增量数组，delta
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int d[8][2] = {
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{1, 2},
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{1, -2},
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{-1, 2},
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{-1, -2},
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{2, 1},
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{2, -1},
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{-2, 1},
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{-2, -1}};
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int main() {
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//读入B点坐标和马的坐标
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cin >> n >> m >> x >> y;
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//马的实际控制范围
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for (int i = 0; i < 8; i++) {
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int tx = x + d[i][0], ty = y + d[i][1];
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if (tx >= 0 && tx <= n && ty >= 0 && ty <= m) ctrl[tx][ty] = 1;
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}
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//马所在的位置你也不能走，也踢你~
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ctrl[x][y] = 1;
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//如果原点在马的控制范围内，那么就是无法出发
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//如果原点不在马的控制范围内，那么就是有一种方法
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if (ctrl[0][0]) f[0][0] = 0;
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else f[0][0] = 1;
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//开始递推
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for (int i = 0; i <= n; i++) //遍历整个棋盘
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for (int j = 0; j <= m; j++) {
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//在马的控制范围内，就放弃掉这个点，路线条数为0，默认就是0，不需要改变,直接 continue即可。
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if (ctrl[i][j])continue;
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//不是第0行，可以从上一行递推过来
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if (i > 0) f[i][j] += f[i - 1][j];
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//不是第0列，可以从上一列递推过来
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if (j > 0) f[i][j] += f[i][j - 1];
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}
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//输出结果
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cout << f[n][m];
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}
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