[BZOJ1879] [Sdoi2009] Bill的挑战 [状态压缩][dp]

[ $\frak{Link}$]

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观察数据范围考虑状态压缩？
考虑把字符串匹配拆成每一位匹配，显然可以每位求与递推，到最后就是匹配状态了。
数组不要开小。

$\frak{ps.}$可以容斥。

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#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cctype>
#include<cstdlib>
#include<queue>
using namespace std;
const int mod = 1000003;
int T, n, k, m;
int a[100][100];
int f[100][50000];
char s[1000][1000];
int main() {
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &n, &k);
		memset(s, 0, sizeof(s));
		for (int i = 1; i <= n; ++i) {
			scanf(" %s", s[i]);
		}
		m = strlen(s[1]);
		for (int i = 0; i < m; ++i) {
			for (int j = 0; j < 26; ++j) {
				a[i][j] = 0;
				for (int k = 1; k <= n; ++k) {
					if (j + 'a' == s[k][i] || s[k][i] == '?') a[i][j] |= 1 << k - 1;
				}
			}
		}
		int up = (1 << n) - 1;
		memset(f, 0, sizeof(f));
		f[0][up] = 1; //* 不能写f[0][0] = 1 
		for (int i = 0; i < m; ++i) {
			for (int j = 0; j <= up; ++j) {
				if (!f[i][j]) continue;
				for (int k = 0; k < 26; ++k) {
					f[i+1][j&a[i][k]] += f[i][j];
					f[i+1][j&a[i][k]] %= mod;
				}
			}
		}
		int ans(0);
		for (int j = 0; j <= up; ++j) { // *
			int cnt(0), t = j;
			while (t) {
				t ^= t & -t;
				++cnt;
			}
			if (cnt == k) ans += f[m][j], ans %= mod;
		}
		printf("%d\n", ans);
	}
    return 0;
}