| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 423652 | jtnydv25 | Shortcut (IOI16_shortcut) | C++17 | 0 ms | 0 KiB |
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 "shortcut.h"
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define all(c) ((c).begin()), ((c).end())
#define sz(x) ((int)(x).size())
#ifdef LOCAL
#include <print.h>
#else
#define trace(...)
#define endl '\n'
#endif
const int N = 1 << 20;
const ll INF = 1e18;
struct Data{
ll a, b;
Data() : a(-INF), b(INF){}
Data(ll a, ll b) : a(a), b(b){}
};
inline void merge(Data & X, const Data & Y){
X.a = max(X.a, Y.a);
X.b = min(X.b,Y.b);
}
Data data[N];
void init(){
for(int i = 0; i < N; i++) data[i] = Data();
}
inline void add(int i, Data d){
i++;
for(; i < N; i += (i & (-i))){
merge(data[i], d);
}
}
Data get(int i){
i++;
Data ret;
for(; i; i -=(i & (-i))) merge(ret, data[i]);
return ret;
}
long long find_shortcut(int n, std::vector<int> l, std::vector<int> d, int c){
vector<ll> p(n);
for(int i = 1; i < n; i++) p[i] = p[i - 1] + l[i - 1];
vector<int> perm1(n), perm2(n);
vector<int> where(n);
iota(all(perm1), 0);
iota(all(perm2), 0);
sort(all(perm1), [&](int i, int j){return d[i] - p[i] > d[j] - p[j];});
sort(all(perm2), [&](int i, int j){return d[i] + p[i] < d[j] + p[j];});
for(int i = 0; i < n; i++) where[perm1[i]] = i;
ll lo = 0, hi = 1000000000LL * 10000000LL;
vector<int> position(n);
while(lo < hi){
ll D = (lo + hi) / 2;
Data d1, d2;
int cur = 0; // < 0
for(int i : perm2){
while(cur < n && d[perm1[cur]] + d[i] + p[i] - p[perm1[cur]] > D) cur++;
position[i] = cur;
}
init();
for(int j = 0; j < n; j++){
Data temp = get(position[j] - 1);
d1.a = max(d1.a, temp.a + d[j] + c - p[j] - D);
d1.b = min(d1.b, temp.b - d[j] - c - p[j] + D);
d2.a = max(d2.a, temp.a + p[j] - D + d[j] + c);
d2.b = min(d2.b, temp.b + p[j] + D - d[j] - c);
add(where[j], Data(p[j] + d[j], p[j] - d[j]));
}
ll L1 = d1.a, R1 = d1.b;
ll L2 = d2.a, R2 = d2.b;
if(L1 > R1 || L2 > R2){
lo = D + 1;
continue;
}
// some u, v with p[u] - p[v] in [L1, R1] and p[u] + p[v] in [L2, R2]
bool ok = false;
for(int v = 1; v < n; v++){
ll L = max(p[v] + L1, L2 - p[v]);
ll R = min(p[v] + R1, R2 - p[v]);
// in L, R?
int u = upper_bound(all(p), L - 1) - p.begin(); // first >= L
if(u >= v) continue;
if(p[u] > R) continue;
ok = true;
break;
}
if(ok) hi = D;
else lo = D + 1;
}
return lo;
}