Submission #1277548

#	Time	Username	Problem	Language	Result	Execution time	Memory
1277548		quanduongxuan12	Crocodile's Underground City (IOI11_crocodile)	C++20	100 / 100	257 ms	47680 KiB

crocodile

#include "crocodile.h"
#include <bits/stdc++.h>
using namespace std;
#define name "crocodile"
#define MAXN 100005
#define pb push_back
#define pf push_front
#define ll long long
#define ii pair<int, int>
#define fs first
#define sc second
#define ill pair<int, ll>
#define lli pair<ll, int>
#define llll pair<ll, ll>
#define all(v) v.begin(),v.end()
#define uni(v) v.erase(unique(all(v)),v.end())
#define bit(n,i) (((n)>>(i))&1)
#define FOR(i,a,b) for (int i=(a),_b=(b); i<=_b; i++)
#define FORD(i,a,b) for (int i=(a),_b=(b); i>=_b; i--)
#define MASK(i) (1LL<<(i))
const ll INF=1e18;
const int MOD=1e9+7;
void add(int &u, int v){
    u+=v;
    if (u>=MOD) u-=MOD;
}
void sub(int &u, int v){
    u-=v;
    if (u<0) u+=MOD;
}
void minimize(int &u, int v){
    u=min(u,v);
}
void maximize(int &u, int v){
    u=max(u,v);
}
long long Rand(long long l, long long r){
    ll tmp=0;
    FOR(i,1,4) tmp=((tmp<<15)^(((1<<15)-1)&rand()));
    return l+tmp%(r-l+1);
}
int n,m,k;
bool spec[MAXN];
vector<ii> adj[MAXN];
ll d[MAXN][2];
void dijkstra(){
    priority_queue<lli,vector<lli>,greater<lli>> pq;
    FOR(i,1,n){
        if (spec[i]){
            d[i][0]=d[i][1]=0;
            pq.push({0LL,i});
        }
        else d[i][0]=d[i][1]=INF;
    }
    while (pq.size()){
        lli tmp=pq.top();
        pq.pop();
        ll w=tmp.fs;
        int u=tmp.sc;
        if (w>d[u][1]) continue;
        for (ii pairs: adj[u]){
            int v=pairs.fs;
            ll wei=pairs.sc;
            if (w+wei<d[v][0]){
                if (d[v][0]<d[v][1]){
                    pq.push({d[v][0],v});
                }
                d[v][1]=d[v][0];
                d[v][0]=w+wei;
            }
            else if (w+wei<d[v][1]){
                d[v][1]=w+wei;
                pq.push({d[v][1],v});
            }
        }
    }
}
int travel_plan(int N, int M, int R[][2], int L[], int K, int P[]){
    n=N;
    m=M;
    k=K;
    FOR(i,0,m-1){
        int x=R[i][0],y=R[i][1];
        ++x;
        ++y;
        int w=L[i];
        adj[x].pb({y,w});
        adj[y].pb({x,w});
    }
    FOR(i,0,k-1) spec[++P[i]]=1;
    dijkstra();
    return d[1][1];
}

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