Tom和Jerry在下棋

题目描述

方法：

状压DP

#include <cstdio>
#define bc(x) (__builtin_popcount(x))
const int mod = 998244353;
const int maxn = 12;
int f[maxn][1 << maxn][maxn * maxn];
int main()
{
    int n, m, p;
    scanf("%d%d%d", &n, &m, &p);
    for (int i = 0; i < 1 << m; ++i) f[1][i][bc(i)] = 1;
    for (int i = 2; i <= n; ++i)
    {
        for (int j = 0; j < 1 << m; ++j)
        {
            for (int k = 0; k < 1 << m; ++k)
            {
                if ((j << 1) & k) continue;
                if (j & (k << 1)) continue;
                for (int l = p; l >= bc(k); --l) f[i][k][l] = (f[i][k][l] + f[i - 1][j][l - bc(k)]) % mod;
            }
        }
    }
    static int ans;
    for (int i = 0; i < 1 << m; ++i) ans = (ans + f[n][i][p]) % mod;
    printf("%d\n", ans);
    return 0;
}