Submission #947581

#	Time	Username	Problem	Language	Result	Execution time	Memory
947581		GrindMachine	Cubeword (CEOI19_cubeword)	C++17	100 / 100	808 ms	13348 KiB

cubeword

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
edi
some codes (to understand the edi idea)

sum the answers over all lengths
how to calculate the ans for a fixed length?

key idea:
when the letters in some vertices are fixed, the #of cubes that satisfy the fixed letters can be quickly counted
which vertices to fix?

fix 4 vertices:
a = front bottom left
b = front top right
c = back bottom right
d = back top left

none of these vertices are adj to each other
look at the 4 unfixed vertices
each one is adjacent to 3 vertices, all of which are fixed
so an unfixed vertex can have any letter, but the words on the 3 edges need to end with the fixed letters
precompute f(a,b,c) = #of ordered triplets of words s.t all 3 words start with the same letter, but end with a,b,c respectively
this can be used to count the #of cubes satisfying the fixed letters in O(1)

for eg, look at the front bottom right vertex
it's adj to a,b,c
so ways to fill these edges = f(a,b,c)
similar argument holds for the other 3 unfixed vertices

at the end, the final formula looks like:
f(a,b,c)*f(a,b,d)*f(b,c,d)*f(a,c,d)

*/

const int MOD = 998244353;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n; cin >> n;
    vector<string> a(n);
    rep(i,n) cin >> a[i];
    rep(i,n){
        auto s = a[i];
        reverse(all(s));
        a.pb(s);
    }

    sort(all(a));
    a.resize(unique(all(a))-a.begin());
    n = sz(a);

    vector<char> b;
    trav(s,a){
        trav(ch,s){
            b.pb(ch);
        }
    }

    sort(all(b));
    b.resize(unique(all(b))-b.begin());
    ll alpha = sz(b);

    ll cnt[15][alpha][alpha];
    memset(cnt,0,sizeof cnt);
    rep(i,n){
        auto &s = a[i];
        ll x = lower_bound(all(b),s[0])-b.begin();
        ll y = lower_bound(all(b),s.back())-b.begin();
        cnt[sz(s)][x][y]++;
    }

    ll f[alpha][alpha][alpha];
    ll ans = 0;

    for(int len = 3; len <= 10; ++len){
        memset(f,0,sizeof f);
        rep(first,alpha){
            rep(x,alpha){
                rep(y,alpha){
                    rep(z,alpha){
                        ll add = cnt[len][first][x]*cnt[len][first][y]*cnt[len][first][z];
                        f[x][y][z] = (f[x][y][z]+add)%MOD;
                    }
                }
            }
        }

        rep(w,alpha){
            rep(x,alpha){
                rep(y,alpha){
                    rep(z,alpha){
                        ll add = f[w][x][y]*f[w][x][z]%MOD*f[x][y][z]%MOD*f[w][y][z];
                        ans = (ans+add)%MOD;
                    }
                }
            }
        }
    }

    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 21 / 21
• Subtask 2: 29 / 29
• Subtask 3: 34 / 34
• Subtask 4: 16 / 16