题解 | #Jobs (Easy Version)#

D Jobs (Easy Version)

bitset邪教，优雅的暴力三维前缀或，完全不需要脑子！来自队友的提交+我修改数组大小。极限卡过空间以及优美的时间（什么？我是全队唯一的单身狗？）。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
uniform_int_distribution<> u(1,400);
const int N = 15, M = 401, mod = 998244353;
bitset<10> s[M][M][M];
int n, m, seed;
LL qmi(LL a, LL b){
    LL res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return res;
}
int main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> m;
    for(int i = 0; i < n; i ++){
        int k;
        cin >> k;
        while(k --){
            int a, b, c;
            cin >> a >> b >> c;
            s[a][b][c][i] = true;
        }
    }
    cin >> seed;
    mt19937 rng(seed);
    for(int j = 1; j <= 400; j ++){
        for(int k = 1; k <= 400; k ++){
            for(int l = 1; l <= 400; l ++){
                s[j][k][l] = (s[j][k][l] | s[j - 1][k - 1][l - 1] | s[j - 1][k - 1][l] | s[j - 1][k][l - 1]
                              | s[j][k - 1][l - 1] | s[j - 1][k][l] | s[j][k - 1][l] | s[j][k][l - 1]);
            }
        }
    }
    int lastans=0, res = 0;
    for (int i=1;i<=m;i++){
        int IQ=(u(rng)^lastans)%400+1;  // The IQ of the i-th friend
        int EQ=(u(rng)^lastans)%400+1;  // The EQ of the i-th friend
        int AQ=(u(rng)^lastans)%400+1;  // The AQ of the i-th friend
        int temp = s[IQ][EQ][AQ].count();
        res = (res + (temp * qmi(seed, m - i)) % mod) % mod;
        lastans = temp; // The answer to the i-th friend
    }
    cout << res << endl;

    return 0;
}