loj#10172 涂抹果酱 (状压DP)

题目：

#10172. 「一本通 5.4 练习 1」涂抹果酱

解析：

三进制的状压DP

经过简单的打表发现，在\(m=5\)时最多有\(48\)种合法状态

然后就向二进制一样枚举当前状态和上一层的状态进行转移就好了

由于第\(k\)行是给定的，所以转移时要特判一下第\(k\)行，并且注意下一\(k=1\)的情况

设\(f[i][j]\)表示第\(i\)行\(j\)状态时的方案数

\(f[i][j] += f[i - 1][k],k是上一行的状态\)

代码：

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e5 + 10;
const int mod = 1e6;

int n, m, k, num, sum = 1, sta, kk, ans;
/*
	sum为状态上节
	sta是第k行状态
	num是合法状态数
*/
int state[N], f[10010][50];
//state记录状态
//f[i][j]表示dp 到第i行j状态的方案数

bool check(int x) {		//判断状态是否合法
	int len = 0;
	int b[10];
	while (x) b[++len] = x % 3, x /= 3;
	if (len < m - 1) return 0;
	for (int i = 2; i <= len; ++i) if (b[i] == b[i - 1]) return 0;
	return 1;
}

bool cmp(int x, int y) {	//判断两个状态是否可以相邻
	for (int i = 1; i <= m; ++i) {
		if ((x % 3) == (y % 3)) return 0;
		x /= 3, y /= 3;
	}
	return 1;
}

signed main() {
	cin >> n >> m >> k;
	for (int i = 1, x; i <= m; ++i) cin >> x, kk = kk * 3 + x - 1;
	for (int i = 1; i <= m; ++i) sum *= 3;
	for (int i = 0; i < sum; ++i) if (check(i)) {
			state[++num] = i;
			if (i == kk) sta = num;
		}
	if (sta == 0) {
		cout << 0;
		return 0;
	}
	for (int i = 1; i <= n; ++i) {
		if (i == k) {
			if (i == 1) f[1][sta] = 1;
			else for (int j = 1; j <= num; ++j)
					if (cmp(state[sta], state[j]))
						(f[i][sta] += f[i - 1][j]) %= mod;
		} else {
			if (i == 1) for (int j = 1; j <= num; ++j) f[i][j] = 1;
			else for (int j = 1; j <= num; ++j)
					for (int pre = 1; pre <= num; ++pre)
						if (cmp(state[j], state[pre]))
							(f[i][j] += f[i - 1][pre]) %= mod;
		}
	}
	for (int i = 1; i <= num; ++i) (ans += f[n][i]) %= mod;
	cout << ans;
}