제출 #446131

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
446131		stat0r1c	악어의 지하 도시 (IOI11_crocodile)	C++11	89 / 100	614 ms	64008 KiB

crocodile

#include <bits/stdc++.h>
#include <crocodile.h>
/* #include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp> */
using namespace std;
// using namespace __gnu_pbds;
using ll = long long;
using ull = unsigned ll;
template<typename T> using vt = vector<T>;
using vi = vt<int>;
using vvi = vt<vi>;
using ii = pair<int, int>;
using vii = vt<ii>;
/* template<typename T>
using indexed_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>; */
/* struct chash {
    static ull hash_f(ull x) {x += 0x9e3779b97f4a7c15; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9; x = (x ^ (x >> 27)) * 0x94d049bb133111eb; return x ^ (x >> 31);}
    size_t operator()(ull x) const {static const ull RANDOM = chrono::steady_clock::now().time_since_epoch().count();return hash_f(x + RANDOM);}
};
struct iihash {
    template <class T1, class T2>
    size_t operator () (const pair<T1,T2> &p) const {auto h1 = hash<T1>{}(p.first);auto h2 = hash<T2>{}(p.second);return h1 ^ h2;}
};
template<class T1, class T2, class T3> using ii_umap = gp_hash_table<pair<T1,T2>,T3,iihash>;
template<class T1, class T2> using f_umap = gp_hash_table<T1,T2,chash>;
template<class T1, class T2> using o_umap = gp_hash_table<T1,T2>; */
#define sqr(x) (x)*(x)
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define debug(x) cerr << #x << " -> " << x << '\n'
#define F first
#define S second
#define mp make_pair
#define pb push_back
#define eb emplace_back
const int inf = 1e9 + 7;
const ll infll = 1e18 + 10;
template <typename T> inline T fgcd(T a, T b) {while(b) swap(b, a %= b); return a;}
ll fpow(ll a, ll b) {ll o = 1; for(;b;b >>= 1) {if(b & 1) o = o * a;a = a * a;} return o;}
ll fpow(ll a, ll b, ll m) {ll o = 1; a %= m; for(;b;b >>= 1) {if(b & 1) o = o * a % m;a = a * a % m;} return o;}
template<class T> void uniq(vt<T> &a) {sort(all(a));a.resize(unique(all(a)) - a.begin());}
/* void setIO(string name = "", string inp = "inp", string out = "out") {
    cin.tie(0)->sync_with_stdio(0);
    if(sz(name)) {
        if(sz(inp)) freopen((name+"."+inp).c_str(), "r", stdin);
        if(sz(out)) freopen((name+"."+out).c_str(), "w", stdout);
    }
} */
using pll = pair<ll, ll>;
int travel_plan(int n, int m, int r[][2], int* l, int k, int* p) {
    vt<vii> a(n);
    for(int i = 0; i < m; i++) {
        a[r[i][0]].eb(l[i],r[i][1]);
        a[r[i][1]].eb(l[i],r[i][0]);
    }
    vt<pll> d(n, {infll, infll});
    priority_queue<pll, vt<pll>, greater<pll>> pq;
    for(int i = 0; i < k; i++) {
        d[p[i]] = {0,0};
        pq.push({0, p[i]});
    }
    while(!pq.empty()) {
        ll du = pq.top().F, u = pq.top().S;
        pq.pop();
        if(du != d[u].S) continue;
        for(ii v : a[u]) {
            if(d[v.S].F > du + v.F) {
                d[v.S].S = d[v.S].F;
                d[v.S].F = du + v.F;
                if(d[v.S].S != infll) pq.push({d[v.S].S, v.S});
            }
            else if(d[v.S].S > du + v.F) pq.push({d[v.S].S = du + v.F, v.S});
        }
    }
    return d[0].S;
}

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