考虑到合法最大状态一定是某个禁用状态 $\{a\}$ 的某个元素减一，即把某个 $a_i$ 变成 $a_{i - 1}$。

因此选择遍历每个禁用状态和状态的每个元素来求解。

由于用了 $map$，复杂度 $O(nm\log n)$（其实原本用的是hash，但是被毒瘤 CF 卡了）

然而不知道为什么 $\text{WA}$ 了 ${\#}44$, 求各路巨神帮忙看看

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>

using namespace std;

using LL = long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;

int read() {
	int s = 0, f = 1;
	char ch = getchar();
	for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48);
	return s * f;
}

int n, m;
vector<int> p[20]; // 记录每个槽可以放的物品
int c[100010][20]; // 禁用状态
int path[20]; // 记录答案
map<vector<int>, bool> Map; // 记录禁用状态

LL get_ans(int *c) {
	LL res = 0;
	for (int i = 1; i <= n; i ++ )
		res = res + p[i][c[i] - 1];
	return res;
}
void copy(int *path, int *c) {
	for (int i = 1; i <= n; i ++ )
		path[i] = c[i];
}

void insert(int *a) { // 插入某个禁用状态
	vector<int> k;
	for (int i = 1; i <= n; i ++ ) k.push_back(a[i]);
	Map[k] = true;
}

bool check(int *a) { // 检查某个状态是否是禁用状态
	vector<int> k;
	for (int i = 1; i <= n; i ++ ) k.push_back(a[i]);
	return Map[k];
}

int main() {
	n = read();
	for (int i = 1; i <= n; i ++ ) {
		int len = read();
		while (len -- ) p[i].push_back(read());
	}
	m = read();
	for (int i = 1; i <= m; i ++ ) {
		for (int j = 1; j <= n; j ++ ) c[i][j] = read();
		insert(c[i]);
	}
	for (int i = 1; i <= n; i ++ ) path[i] = p[i].size();
	if (!check(path)) {
		for (int i = 1; i <= n; i ++ ) printf("%d ", path[i]);
		return 0;
	}
	
	LL ans = -100;
	for (int i = 1; i <= m; i ++ ) { // 遍历每个禁用状态
		for (int j = 1; j <= n; j ++ ) { // 遍历禁用状态的每个元素
			if (c[i][j] == 1) continue;
			c[i][j] -- ;
			if (check(c[i])) continue;
			if (get_ans(c[i]) > ans) {
				ans = get_ans(c[i]);
				copy(path, c[i]);
			}
			c[i][j] ++ ;
		}
	}
	
	for (int i = 1; i <= n; i ++ ) printf("%d ", path[i]);
	return 0;
}