제출 #1185945

#제출 시각아이디문제언어결과실행 시간메모리
1185945trvhungCrocodile's Underground City (IOI11_crocodile)C++20
100 / 100
256 ms47784 KiB
#include "crocodile.h" #include <bits/stdc++.h> // #include <ext/rope> // #include <ext/pb_ds/assoc_container.hpp> // using namespace __gnu_pbds; // using namespace __gnu_cxx; using namespace std; // #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> #define ll long long #define ull unsigned long long #define ld long double #define pb push_back #define bit(mask, i) ((mask >> i) & 1) #define el '\n' #define F first #define S second template <class X, class Y> bool maximize(X &x, const Y &y) { return (x < y ? x = y, 1 : 0); } template <class X, class Y> bool minimize(X &x, const Y &y) { return (x > y ? x = y, 1 : 0); } const int INF = 1e9; const ll LINF = 1e18; const int MOD = 1e9 + 7; const int MULTI = 0; const ld eps = 1e-9; const int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0}; // R D L U const int ddx[4] = {-1, 1, 1, -1}, ddy[4] = {1, 1, -1, -1}; // UR DR DL UL const char cx[4] = {'R', 'D', 'L', 'U'}; const ll base = 31; const int nMOD = 2; const ll mods[] = {(ll)1e9 + 10777, (ll)1e9 + 19777, (ll)1e9 + 3, (ll)1e9 + 3777}; const int maxn = 1e5 + 5; int n, m, k, res; vector<pair<int, int>> adj[maxn]; bool is_exit[maxn]; ll dist[2][maxn]; void opt(int u, ll d) { if (d < dist[0][u]) dist[1][u] = dist[0][u], dist[0][u] = d; else if (d < dist[1][u]) dist[1][u] = d; } void solve() { priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq; for (int i = 1; i <= n; ++i) if (is_exit[i]) { pq.push(make_pair(0, i)); dist[0][i] = dist[1][i] = 0; } else dist[0][i] = dist[1][i] = LINF; while (!pq.empty()) { auto top = pq.top(); pq.pop(); ll d = top.F; int u = top.S; if (dist[1][u] != d) continue; for (auto g: adj[u]) { int v = g.F, w = g.S; ll old = dist[1][v]; opt(v, d + w); if (dist[1][v] != old) pq.push(make_pair(dist[1][v], v)); } } res = dist[1][1]; } int travel_plan(int N, int M, int R[][2], int L[], int K, int P[]) { n = N; m = M; for (int i = 0; i < m; ++i) { int u = R[i][0] + 1, v = R[i][1] + 1, w = L[i]; adj[u].push_back(make_pair(v, w)); adj[v].push_back(make_pair(u, w)); } k = K; for (int i = 0; i < k; ++i) is_exit[P[i] + 1] = true; solve(); return res; }
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