This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#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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