이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "sequence.h"
#include "bits/stdc++.h"
using namespace std;
#ifndef EVAL
#include "grader.cpp"
#endif
#define ar array
typedef long long ll;
//~ #define int ll
const int INF = 1e9;
const int N = 5e5 + 5;
int A_[N], B_[N];
struct ST{
int min[N << 2], max[N << 2], p[N << 2];
ST(){
memset(min, 0, sizeof min);
memset(max, 0, sizeof max);
memset(p, 0, sizeof p);
}
void push(int x, int lx, int rx){
if(lx == rx) return;
min[x << 1] += p[x], max[x << 1] += p[x], p[x << 1] += p[x];
min[x << 1 | 1] += p[x], max[x << 1 | 1] += p[x], p[x << 1 | 1] += p[x];
p[x] = 0;
}
void add(int l, int r, int v, int lx, int rx, int x){
if(lx > r || rx < l) return;
if(lx >= l && rx <= r){
min[x] += v, max[x] += v, p[x] += v;
return;
}
int m = (lx + rx) >> 1;
push(x, lx, rx);
add(l, r, v, lx, m, x << 1), add(l, r, v, m + 1, rx, x << 1 | 1);
min[x] = ::min(min[x << 1], min[x << 1 | 1]);
max[x] = ::max(max[x << 1], max[x << 1 | 1]);
}
void add(int l, int r, int v){
add(l, r, v, 0, N, 1);
}
int Max(int l, int r, int lx, int rx, int x){
if(lx > r || rx < l) return -INF;
if(lx >= l && rx <= r) return max[x];
int m = (lx + rx) >> 1;
push(x, lx, rx);
return ::max(Max(l, r, lx, m, x << 1), Max(l, r, m + 1, rx, x << 1 | 1));
}
int Max(int l, int r){
return Max(l, r, 0, N, 1);
}
int Min(int l, int r, int lx, int rx, int x){
if(lx > r || rx < l) return INF;
if(lx >= l && rx <= r) return min[x];
int m = (lx + rx) >> 1;
push(x, lx, rx);
return ::min(Min(l, r, lx, m, x << 1), Min(l, r, m + 1, rx, x << 1 | 1));
}
int Min(int l, int r){
return Min(l, r, 0, N, 1);
}
};
int sequence(int n, vector<int> a){
a.insert(a.begin(), 0);
ST A, B;
vector<int> p(n);
iota(p.begin(), p.end(), 1);
sort(p.begin(), p.end(), [&](int i, int j){
if(a[i] != a[j]) return a[i] < a[j];
return i < j;
});
for(int i=1;i<=n;i++){
B.add(i, i, i);
A.add(i, i, -i);
}
int res = 0;
for(int i=0,j=0;i<n;){
while(j < n && a[p[i]] == a[p[j]]){
A.add(p[j], n, 2);
j++;
}
for(int k=0;i + k<j;k++){
A_[k] = A.Min(0, p[i + k] - 1);
B_[k] = B.Min(0, p[i + k] - 1);
int MxA = A.Max(p[i + k], n), MxB = B.Max(p[i + k], n);
int l = 0, r = k;
while(l < r){
int m = (l + r) >> 1;
if(A_[m] <= MxA && B_[m] <= MxB) r = m;
else l = m + 1;
}
res = max(res, k - l + 1);
}
while(i < j){
B.add(p[i], n, -2);
i++;
}
}
return res;
}
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