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>
using namespace std;
const int oo = 1e9;
struct Tree {
typedef pair<int, int> T;
static constexpr T unit = {-oo, -oo};
T f(T a, T b) { return max(a, b); }
vector<T> s;
int n;
Tree(int nn) : s(2 * nn, unit), n(nn) {}
void update(int pos, T val)
{
for(s[pos += n] = val; pos /= 2;) {
s[pos] = f(s[pos * 2], s[pos * 2 + 1]);
}
}
T query(int b, int e)
{
T ra = unit, rb = unit;
for(b += n, e += n + 1; b < e; b /= 2, e /= 2) {
if(b % 2) ra = f(ra, s[b++]);
if(e % 2) rb = f(s[--e], rb);
}
return f(ra, rb);
}
};
struct treatment {
int t, l, r, c;
pair<int, int> pocz, kon;
void read()
{
cin >> t >> l >> r >> c;
r++;
pocz.first = l + t;
pocz.second = t - l;
kon.first = r + t;
kon.second = t - r;
}
bool operator<(const treatment &he) const
{
return pocz.first < he.pocz.first;
}
};
int main()
{
cin.tie(nullptr)->sync_with_stdio(0);
int n, m;
cin >> n >> m;
vector<treatment> they(m + 1);
for(int i = 1; i <= m; i++) {
they[i].read();
}
sort(they.begin() + 1, they.end());
Tree tree(m + 1);
#define T pair<long long, int>
priority_queue<T, vector<T>, greater<T>> q;
const long long ool = 1e18 + 5;
vector<long long> d(m + 1);
for(int i = 1; i <= m; i++) {
if(they[i].l == 1) {
d[i] = they[i].c;
q.push({d[i], i});
}
else {
d[i] = ool;
tree.update(i, {they[i].pocz.second, i});
}
}
while(!q.empty()) {
pair<long long, int> t = q.top();
q.pop();
while(true) {
int lo = 1, hi = m, ans = 0;
while(lo <= hi) {
int mid = (lo + hi) / 2;
if(they[t.second].kon.first < they[mid].pocz.first) {
hi = mid - 1;
}
else {
lo = mid + 1;
ans = mid;
}
}
pair<int, int> maybe = tree.query(1, ans);
if(maybe.first < they[t.second].kon.second) {
break;
}
assert(d[maybe.second] > they[maybe.second].c + t.first);
q.emplace(d[maybe.second] = they[maybe.second].c + t.first, maybe.second);
tree.update(maybe.second, {-oo, -oo});
}
}
long long ans = ool;
for(int i = 1; i <= m; i++) {
if(they[i].r == n + 1) {
ans = min(ans, d[i]);
}
}
cout << (ans == ool ? -1 : ans) << '\n';
}
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