[USACO06NOV] Corn Fields G

[USACO06NOV] Corn Fields G

状压 \(DP\) 板子题, 也是我的第一道状压 \(DP\) .

设 \(f[i][j]\) 为第 \(i\) 行状态为 \(j\) 时的方案数, 则上下两行没有相邻的用 \(j \& k == 0\) 来判, 同一行没有相邻的用 \(k << 1 \& k == 0 \&\& k >> 1 \& k == 0\) 来判.

\(code:\)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
const int mod = 1e8;
int n, m, f[15][1 << 12], s[15];
int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            s[i] = (s[i] << 1) + read();
        }
    }
    f[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < (1 << m); j++) {
            for (int k = 0; k < (1 << m); k++) {
                if (!(k & j) && (s[i] & k) == k && !(k << 1 & k) && !(k >> 1 & k)) {
                    f[i][k] += f[i - 1][j];
                    if (f[i][k] >= mod) f[i][k] -= mod;
                }
            }
        }
    }
    int ans = 0;
    for (int i = 0; i < (1 << m); i++) {
        ans += f[n][i];
        if (ans >= mod) ans -= mod;
    }
    printf("%d", ans);
    return 0;
}