Submission #719921

#TimeUsernameProblemLanguageResultExecution timeMemory
719921joelgun14Jakarta Skyscrapers (APIO15_skyscraper)C++17
100 / 100
794 ms70352 KiB
#include <bits/stdc++.h> #pragma GCC optimize("Ofast") #define mp make_pair #define fi first #define lb lower_bound #define se second #define endl "\n" using namespace std; struct custom_hash { static uint64_t splitmix64(uint64_t x) { // http://xorshift.di.unimi.it/splitmix64.c x += 0x9e3779b97f4a7c15; x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9; x = (x ^ (x >> 27)) * 0x94d049bb133111eb; return x ^ (x >> 31); } size_t operator()(uint64_t x) const { static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count(); return splitmix64(x + FIXED_RANDOM); } }; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n, m; cin >> n >> m; int blk = 50; vector<int> weights[n]; int init = -1, initp = -1, target = -1; for(int i = 1; i <= m; ++i) { int b, p; cin >> b >> p; if(i == 1) init = b, initp = p; else if(i == 2) target = b; // b -> pos // p -> multiple weights[b].push_back(p); } // gaperlu cek semua cukup cek yg ada weights // kalo udh proc weight jg gperlu // kalo <= blk maka memo tiap pair int modulo dalam vector // kalo > blk maka cek manual // doge cmn boleh gerak dr ori node (gaboleh meet di tengah) // hence cukup dijkstra dengan 1 origin, terus klo intersect lainnya bs langsung dijkstra aja // tiap node ada theoretically n^2 dijkstra state yg mgkn? // tp yg high modulo itu sparse, jadi bs simpan manual pakai sorted vector/semacamnya priority_queue<pair<int, pair<int, pair<int, int>>>, vector<pair<int, pair<int, pair<int, int>>>>, greater<pair<int, pair<int, pair<int, int>>>>> pq; // fi -> dist // se.fi -> idx // se.se.fi -> current p // se.se.se -> tanda dr atas/bawah // guna -> reduce double counting //cout << init << " " << initp << endl; pq.push(mp(0, mp(init, mp(initp, -1)))); int d[n + 1]; map<int, bool> vis[n]; memset(d, -1, sizeof(d)); // cek ada yg bs sampai ke target atau tidak, kalo tidak langsung output false // kalo iya, maka ada pembuktian min path tidak sepanjang itu? while(pq.size()) { int dist = pq.top().fi, idx = pq.top().se.fi, curp = pq.top().se.se.fi, tanda = pq.top().se.se.se; pq.pop(); //cout << dist << " " << idx << " " << curp << endl; if(vis[idx][curp]) continue; if(d[idx] == -1) d[idx] = dist; d[idx] = min(d[idx], dist); for(auto i : weights[idx]) { if(!vis[idx][i] && i != curp) { if(i > blk) { int tmp = idx + i; while(tmp < n && !weights[tmp].size()) tmp += i; if(tmp < n && !vis[tmp][i]) pq.push(mp(dist + (tmp - idx) / i, mp(tmp, mp(i, 0)))); tmp = idx - i; while(tmp >= 0 && !weights[tmp].size()) tmp -= i; if(tmp >= 0 && !vis[tmp][i]) pq.push(mp(dist + (idx - tmp) / i, mp(tmp, mp(i, 1)))); } else { if(idx + i < n && !vis[idx + i][i]) pq.push(mp(dist + 1, mp(idx + i, mp(i, 0)))); if(idx - i >= 0 && !vis[idx - i][i]) pq.push(mp(dist + 1, mp(idx - i, mp(i, 1)))); } vis[idx][i] = 1; } } weights[idx].clear(); //cout << idx << " " << curp << endl; if(!vis[idx][curp]) { vis[idx][curp] = 1; //cout << dist << " " << idx << " " << curp << endl; if(curp > blk) { int tmp = idx + curp; if(tanda != 1) { while(tmp < n && !weights[tmp].size()) tmp += curp; if(tmp < n && !vis[tmp][curp]) pq.push(mp(dist + (tmp - idx) / curp, mp(tmp, mp(curp, 0)))); } if(tanda != 0) { tmp = idx - curp; while(tmp >= 0 && !weights[tmp].size()) tmp -= curp; if(tmp >= 0 && !vis[tmp][curp]) pq.push(mp(dist + (idx - tmp) / curp, mp(tmp, mp(curp, 1)))); } } else { if(idx + curp < n && !vis[idx + curp][curp]) pq.push(mp(dist + 1, mp(idx + curp, mp(curp, 0)))); if(idx - curp >= 0 && !vis[idx - curp][curp]) pq.push(mp(dist + 1, mp(idx - curp, mp(curp, 1)))); } } if(idx == target) break; } cout << d[target] << endl; }
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