2019 CCF CSP-S2.Emiya 家今天的饭

题目

1155. Emiya 家今天的饭

思路

容斥原理, $dp+dp$, 不是$dpofdp$
核心方法总得方案数减去不合法的方案数, 首先考虑前两个约束, 状态表示$f[i][j]$表示前$i$个烹饪方法,使用了$j$个食材的所有方案, 类似于背包问题预处理

	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= i; ++j) {
			f[i][j] = f[i - 1][j];
			if (j) f[i][j] = (f[i][j] + (LL) f[i - 1][j - 1] * s[i] % MOD) % MOD;
		}
	}

	int res = 0;
	for (int i = 1; i <= n; ++i) res = (res + f[n][i]) % MOD;

再进行$dp$, 定义$g[i][d]$为考虑前$i$个烹饪方式, 对于第$k$个食材, 第$k$个食材出现的次数减去其他食材出现的次数为$d$的所有方案

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;
const int N = 110, M = 2010, MOD = 998244353;

int n, m;
LL w[N][M], s[N];
// f[i][j]前i个烹饪方式, 使用了j个食材的所有方案, g[i][d]表示前i个烹饪方式对于食材k来说所有菜中k出现的次数减去其他食材出现的次数是d的所有方案
int f[N][M], g[N][N << 1];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n >> m;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			cin >> w[i][j];
			s[i] = ((LL) s[i] + w[i][j]) % MOD;
		}
	}

	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= i; ++j) {
			f[i][j] = f[i - 1][j];
			if (j) f[i][j] = (f[i][j] + (LL) f[i - 1][j - 1] * s[i] % MOD) % MOD;
		}
	}

	int res = 0;
	for (int i = 1; i <= n; ++i) res = (res + f[n][i]) % MOD;

	for (int k = 1; k <= m; ++k) {
		memset(g, 0, sizeof g);
		g[0][N] = 1;
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j + 1 < 2 * N; ++j) {
				// 不选择当前烹饪方法
				g[i][j] = g[i - 1][j];
				// 选择食材k
				g[i][j] = (g[i - 1][j] + (LL) g[i - 1][j - 1] * w[i][k]) % MOD;
				// 不选择食材k
				g[i][j] = (g[i][j] + (LL) g[i - 1][j + 1] * (s[i] - w[i][k])) % MOD;
			}
		}

		for (int i = 1; i <= n; ++i) res = (res - g[n][N + i]) % MOD;
	}

	cout << (res + MOD) % MOD << "\n";
	return 0;
}