Submission #414070

#	Time	Username	Problem	Language	Result	Execution time	Memory
414070		HediChehaidar	Crocodile's Underground City (IOI11_crocodile)	C++17	46 / 100	4 ms	3276 KiB

crocodile

/*
ID: hedichehaidar
TASK: photo
LANG: C++11
*/
#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (PGCD)
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);} // least common multiple (PPCM)
#define ss second
#define ff first
#define all(x) (x).begin() , (x).end()
#define pb push_back
#define vi vector<int>
#define vii vector<pair<int,int>>
#define vl vector<ll>
#define vll vector<pair<ll,ll>>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd  pair<double,double>
#define vdd  vector<pdd>
#define dte  tuple<double , double , double>
using namespace std;
const int INF = 1000*1000*1000; // 1 e 9
const int MOD = INF + 7;
const double EPS = 0.000000001; // 1 e -9
const ll inf = (ll)1e18;

int n;
vii adj[(int)1e5 + 10];
bool ok[(int)1e5 + 10];
bool vis[(int)1e5 + 10];
int dp[(int)1e5 + 10];

int solve(int pos){
    if(ok[pos]) return 0;
    //if(dp[pos] != -1) return dp[pos];
    if(vis[pos]) return MOD;
    vis[pos] = true;
    vi res;
    for(auto c : adj[pos]){
        res.pb(solve(c.ff) + c.ss);
    }
    sort(all(res));
   /* return dp[pos] = */return res[1];
}
int travel_plan(int N , int M , int R[][2] , int L[] , int K , int P[]){
    n = N;
    for(int i = 0 ; i < M ; i++){
        int a = R[i][0] , b = R[i][1] , w = L[i];
        adj[a].pb({b , w});
        adj[b].pb({a , w});
    }
    for(int i = 0 ; i < K; i++) ok[P[i]] = true;
    memset(dp , -1 , sizeof dp);
    return solve(0);
}
/*int main() {
    //ifstream fin ("race.in");
    //ofstream fout ("race.out");
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int N , M ; cin>>N>>M;
    int R[M][2] , L[M] , K ; cin>>K;
    for(int i = 0 ; i < M ; i++){
        cin>>R[i][0]>>R[i][1]>>L[i];
    }
    int P[K]; for(int i = 0 ; i < K ; i++) cin>>P[i];
    cout << travel_plan(N , M , R , L , K , P);
    return 0;
}*/
/*
5 7 2
0 2 4
0 3 3
3 2 2
2 1 10
0 1 100
0 4 7
3 4 9
1 3
*/
/*
    Think of : BS / DFS / BFS / SSSP / SCC / MSP / MAX FLOW / TOPSORT / LCA / MATRIX / DP(bitmask) / 2 POINTERS / SEG TREE / MATH / UN FIND / MO
    Read the statement CAREFULLY !!
    Make a GREADY APPROACH !!!! (start from highest / lowest)
    Make your own TESTS !!
    Be careful from CORNER CASES !
*/

• Subtask 1: 46 / 46
• Subtask 2: 0 / 43
• Subtask 3: 0 / 11