Submission #1282548

#	Time	Username	Problem	Language	Result	Execution time	Memory
1282548		ender_shayan	Paths (BOI18_paths)	C++20	53 / 100	266 ms	56608 KiB

paths

#include <bits/stdc++.h>

using namespace std;

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

typedef long long ll;
typedef long double	ld;
typedef pair<int, int>	pii  ;
typedef pair<ll, ll>	pll  ;
typedef vector<pii>     vii  ;
typedef vector<int>     veci ;
typedef vector<pll>     vll  ;
typedef vector<ll>      vecll;

// find_by_order             order_of_key

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define F		        first
#define S		        second
#define pb		        push_back
#define endl            '\n'
#define Mp		        make_pair
#define all(x)          x.begin(), x.end()
#define debug(x)        cerr << #x << " = " << x << endl
#define set_dec(x)	    cout << fixed << setprecision(x);
#define fast_io         ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io         freopen("in.txt" , "r" , stdin) ; freopen("out.txt" , "w" , stdout);
#define lb              lower_bound
#define ub              upper_bound
#define for1(n)         for(int i=1;i<=n;i++)
#define for0(n)         for(int i=0;i<n;i++)
#define forn(n)         for(int i=n;i>0;i--)
#define pq              priority_queue <pii, vector<pii>, greater<pii>>


const ll mod = 1e9+7 ;// 998244353 ;// 1e9+9;

ll inf=1e18;
const int N=1e5+100,L=21,bs=701;
int A[N],B[N],C[N],D[N],E[N],n,m,k,q,dist[N],vis[N];
vector<int>g[N];
ll ans,dp[N][32],dp2[N][32];
int main(){
    fast_io
    cin>>n>>m>>k;
    for1(n){cin>>A[i];A[i]--;}
    for1(m){
        int u,v;cin>>u>>v;
        g[u].pb(v);
        g[v].pb(u);
    }
    for1(n)dp[i][1<<A[i]]++;
    for(int _=0;_<k;_++){
        for1(n){
            for(int msk=0;msk<(1<<k);msk++){
                for(int u:g[i])if((msk>>A[u]&1)==0)
                    dp2[u][msk|(1<<A[u])]+=dp[i][msk];
            }
        }
        for1(n){
            for(int msk=0;msk<(1<<k);msk++){
                dp[i][msk]=dp2[i][msk];
                dp2[i][msk]=0;
                ans+=dp[i][msk];
            }
        }
    }
    cout<<ans<<endl;




}

• Subtask 1: 23 / 23
• Subtask 2: 0 / 20
• Subtask 3: 0 / 27
• Subtask 4: 30 / 30