洛谷P1002过河卒-题解

原题：

 

 思路：

显然是计数DP

又是在网格上，又是只能走下和右，思路是很显然的

想要到达这个点，只能由这个点上方和左方的点转移来

方程也很明显了：

f[i][j]=f[i-1][j]+f[i][j+1];

注意特判被马占的点就可以了

代码：

#include <bits/stdc++.h>
using namespace std;
int n, m, x, y;
unsigned long long f[25][25];
int book[25][25];
int fx[] = {0, -2, -1, 1, 2, 2, 1, -1, -2};
int fy[] = {0, 1, 2, 2, 1, -1, -2, -2, -1};
int main()
{
    scanf("%d %d %d %d", &n, &m, &x, &y);
    n++;
    m++;
    x++;
    y++;
    book[x][y] = 1;
    f[1][1] = 1;
    for (int i = 1; i <= 8; i++)
        book[x + fx[i]][y + fy[i]] = 1;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (book[i][j] == 0)
                f[i][j] = max(f[i - 1][j] + f[i][j - 1], f[i][j]);
    printf("%llu\n", f[n][m]);
    return 0;
}