Submission #992070

#	Time	Username	Problem	Language	Result	Execution time	Memory
992070		parlimoos	Amusement Park (CEOI19_amusementpark)	C++14	0 / 100	1 ms	2396 KiB

amusementpark

//Be Name KHODA
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
#define pb push_back
#define pp pop_back
#define cl clear
#define bg begin
#define lb lower_bound
#define ub upper_bound
#define arr(x) array<int , x>
#define endl '\n'

const int MOD = 998244353;

int n , m;
int g[18][18] , inf[18];
int dp[18][(1 << 18)][2];

int Sum(int a , int b){
    int res = a + b;
    if(res > MOD) return res - MOD;
    if(res < 0) return res + MOD;
    return res;
}
void Add(int &a , int b){
    a += b;
    if(a > MOD) a -= MOD;
    if(a < 0) a += MOD;
}
int Mul(int a , int b){
    return (1ll * a * b) % MOD;
}
void f(){
    for(int b = 0 ; b < n ; b++) dp[b][0][0] = 1;
    for(int msk = 1 ; msk < (1 << n) ; msk++){
        for(int v = 0 ; v < n ; v++){
            if(!((msk >> v) & 1)) continue;
            inf[v] = 0;
            for(int u = 0 ; u < v ; u++) if(((msk >> u) & 1) and g[u][v] == 1) inf[v]++;
        }
        if(msk == (1 << n) - 1){
            for(int v = 0 ; v < n ; v++){
                Add(dp[0][msk][0] , dp[v][msk ^ (1 << v)][0]);
                Add(dp[0][msk][1] , Sum(Mul(inf[v] , dp[v][msk ^ (1 << v)][0]) , dp[v][msk ^ (1 << v)][1])); 
            }
        }else{
            for(int b = 0 ; b < n ; b++){
                if(((msk >> b) & 1)) continue;
                for(int v = 0 ; v < n ; v++){
                    if(!((msk >> v) & 1) or (v <= b and g[v][b] == 0)) continue;
                    Add(dp[b][msk][0] , dp[v][msk ^ (1 << v)][0]);
                    Add(dp[b][msk][1] , Sum(Mul(inf[v] , dp[v][msk ^ (1 << v)][0]) , dp[v][msk ^ (1 << v)][1]));
                }
            }
        }
    }
}

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

    cin >> n >> m;
    for(int i = 0 ; i < m ; i++){
        int v , u;
        cin >> v >> u;
        v-- , u--;
        g[v][u] = 1 , g[u][v] = -1;
    }
    f();
    cout << dp[0][(1 << n) - 1][1];
}

• Subtask 1: 0 / 7
• Subtask 2: 0 / 12
• Subtask 3: 0 / 23
• Subtask 4: 0 / 21
• Subtask 5: 0 / 37