【LOJ3042】「ZJOI2019」麻将

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_39972971/article/details/88981526

【题目链接】

• 点击打开链接

【思路要点】

• 首先考虑如何判断一副牌是否胡牌。
• 我们可以用动态规划来解决该问题，可以参考 【CodeForces】CodeForces Global Round 1 题解 一文中 $D$ 题的解法，我们可以计算一副牌最多可以组成的面子数，再记一维状态表示是否组成过对子即可。
• 回到原题，我们将上述 $dp$ 的状态、数值全部压入状态，看做当前 $dp$ 的状态，经过搜索，这样的状态至多有 $3956$ 个。
• 考虑对于每个权值 $i$ ，计算所有大小为 $13+i$ 的牌集中不能胡牌的集合数 $X$ 和总集合数 $Y$ ，那么 $\frac{X}{Y}$ 就是权值大于 $i$ 的概率， $\sum\frac{X}{Y}$ 即为权值的期望。
• 记 $dp_{i,j,S}$ 表示当前处理了 $1\sim i$ ，共选定了 $j$ 张牌，胡牌状态为 $S$ 的集合数，转移只需枚举 $i+1$ 的张数即可。
• 时间复杂度 $O(N^2S)$ ，其中 $S=3956$ 。
• 提示：七对子要求对子的大小两两不同。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 105;
const int MAXM = 405;
const int MAXS = 4e3 + 5;
const int P = 998244353;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
struct state {int dp[3][3]; };
bool operator < (state a, state b) {
	for (int i = 0; i <= 2; i++)
	for (int j = 0; j <= 2; j++) {
		if (a.dp[i][j] < b.dp[i][j]) return true;
		if (a.dp[i][j] > b.dp[i][j]) return false;
	}
	return false;
}
int statecnt; map <state, int> mp;
state cipher() {
	state res;
	for (int i = 0; i <= 2; i++)
	for (int j = 0; j <= 2; j++)
		res.dp[i][j] = -1;
	return res;
}
state starter() {
	state res = cipher();
	res.dp[0][0] = 0;
	return res;
}
state operator + (state a, state b) {
	state res;
	for (int i = 0; i <= 2; i++)
	for (int j = 0; j <= 2; j++)
		res.dp[i][j] = max(a.dp[i][j], b.dp[i][j]);
	return res;
}
state operator + (state now, int num) {
	state res = cipher();
	for (int i = 0; i <= 2 && i <= num; i++)
	for (int j = 0; j <= 2 && i + j <= num; j++) {
		if (now.dp[i][j] == -1) continue;
		int tmp = now.dp[i][j];
		for (int k = 0; k <= 2 && i + j + k <= num; k++)
			chkmax(res.dp[j][k], min(tmp + i + (num - i - j - k) / 3 , 4));
	}
	return res;
}
void dfs(state now) {
	if (mp.count(now)) return;
	mp[now] = ++statecnt;
	for (int i = 0; i <= 4; i++)
		dfs(now + i);
}
typedef pair <pair <state, state>, int> mahjong;
//first : without pair, second : with pair
bool goal[MAXS];
mahjong states[MAXS];
map <mahjong, int> home;
int tots, trans[MAXS][5];
pair <state, state> operator + (pair <state, state> now, int num) {
	if (num >= 2) return make_pair(now.first + num, (now.second + num) + (now.first + (num - 2)));
	else return make_pair(now.first + num, now.second + num);
}
mahjong operator + (mahjong now, int num) {
	return make_pair(now.first + num, min(now.second + (num >= 2), 7));
}
void dfm(mahjong now) {
	if (home.count(now)) return;
	home[now] = ++tots;
	states[tots] = now;
	for (int i = 0; i <= 4; i++)
		dfm(now + i);
}
mahjong inception() {
	return make_pair(make_pair(starter(), cipher()), 0);
}
bool finish(mahjong now) {
	if (now.second >= 7) return true;
	for (int i = 0; i <= 2; i++)
	for (int j = 0; j <= 2; j++)
		if (now.first.second.dp[i][j] >= 4) return true;
	return false;
}
void initstates() {
	dfs(starter());
	dfm(inception());
	for (int i = 1; i <= tots; i++) {
		goal[i] = finish(states[i]);
		for (int j = 0; j <= 4; j++)
			trans[i][j] = home[states[i] + j];
	}
}
int power(int x, int y) {
	if (y == 0) return 1;
	int tmp = power(x, y / 2);
	if (y % 2 == 0) return 1ll * tmp * tmp % P;
	else return 1ll * tmp * tmp % P * x % P;
}
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int n, dp[MAXN][MAXM][MAXS];
int binom[5][5], used[MAXN];
int main() {
	initstates(), read(n);
	for (int i = 1; i <= 13; i++) {
		int x, y; read(x), read(y);
		used[x]++;
	}
	for (int i = 0; i <= 4; i++) {
		binom[i][0] = 1;
		for (int j = 1; j <= i; j++)
			binom[i][j] = binom[i - 1][j - 1] + binom[i - 1][j];
	}
	dp[0][0][1] = 1;
	for (int i = 0; i <= n - 1; i++)
	for (int j = 0; j <= 4 * i; j++)
	for (int k = 1; k <= tots; k++) {
		int tmp = dp[i][j][k];
		for (int t = used[i + 1]; t <= 4; t++)
			update(dp[i + 1][j + t][trans[k][t]], 1ll * tmp * binom[4 - used[i + 1]][t - used[i + 1]] % P);
	}
	int ans = 0;
	for (int i = 13; i <= 4 * n; i++) {
		int sum = 0, inv = 0;
		for (int j = 1; j <= tots; j++) {
			update(sum, dp[n][i][j]);
			if (!goal[j]) update(inv, dp[n][i][j]);
		}
		update(ans, 1ll * inv * power(sum, P - 2) % P);
	}
	writeln(ans);
	return 0;
}