이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define ff first
#define ss second
#define pb push_back
#define SZ(x) ((int)(x).size())
#define all(x) x.begin(), x.end()
#define debug(x) cout << #x << ": " << x << " "
#define nl cout << "\n"
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 2e6+100;
int N, D, T, a[MAXN];
vector<int> g[MAXN];
void compute_time(){
stack<int> st;
rep(i, 0, N - 1){
while(!st.empty()){
int j = st.top();
if(a[j] + i - j > T) st.pop();
else break;
}
if(a[i] <= T){
g[i].pb(i);
} else{
if(!st.empty()) g[st.top()].pb(i);
}
st.push(i);
}
}
struct Node{
pii dp = {0, 0}; int lazy = 0;
};
class SegmentTree{
public:
vector<Node> st;
void build(int n){
st = vector<Node>(4 * n);
}
void push(int i){
if(st[i].lazy){
int cl = 2 * i + 1, cr = 2 * i + 2;
st[cl].dp.ff += st[i].lazy, st[cl].lazy += st[i].lazy;
st[cr].dp.ff += st[i].lazy, st[cr].lazy += st[i].lazy;
st[i].lazy = 0;
}
}
void merge(int i){
int cl = 2 * i + 1, cr = 2 * i + 2;
if(st[cl].dp < st[cr].dp){
st[i].dp = st[cl].dp;
} else{
st[i].dp = st[cr].dp;
}
}
void add(int i, int l, int r, int ql, int qr, int k){
if(ql <= l && r <= qr){
st[i].dp.ff += k, st[i].lazy += k; return;
}
if(qr < l || ql > r) return;
push(i);
int m = (l + r) / 2;
add(2 * i + 1, l, m, ql, qr, k);
add(2 * i + 2, m + 1, r, ql, qr, k);
merge(i);
}
void set(int i, int l, int r, int p, pii val){
if(l == r){
st[i].dp = val; st[i].lazy = 0; return;
} else{
push(i);
int m = (l + r) / 2;
if(p <= m) set(2 * i + 1, l, m, p, val);
else set(2 * i + 2, m + 1, r, p, val);
merge(i);
}
}
pii query(int i, int l, int r, int ql, int qr){
if(ql <= l && r <= qr){
return st[i].dp;
}
if(qr < l || ql > r) return {INT_MAX, 0};
push(i);
int m = (l + r) / 2;
pii ans1 = query(2 * i + 1, l, m, ql, qr);
pii ans2 = query(2 * i + 2, m + 1, r, ql, qr);
if(ans1 < ans2) return ans1;
else return ans2;
}
} st;
pii compute_dp(int penalty){
st.build(N + 1);
pii ans; int dpN = 0;
per(i, N - 1, 0){
for(int j : g[i]){
st.add(0, 0, N, j + 1, N, +1);
dpN++;
}
pii val = st.query(0, 0, N, i + 1, N);
val.ff += penalty, val.ss += 1;
pii val2 = {dpN, 0};
if(val > val2) val = val2;
st.set(0, 0, N, i, val);
if(i == 0) ans = val;
}
return ans;
}
void solve(){
cin >> N >> D >> T;
rep(i, 0, N - 1) cin >> a[i];
compute_time();
int l = -1, r = N + 1;
while(r - l > 1){
int m = (l + r) / 2;
pii ans = compute_dp(m);
if(ans.ss <= D) r = m;
else l = m;
}
pii ans = compute_dp(r);
cout << ans.ff - D * r;
}
int main(){
ios_base::sync_with_stdio(false), cin.tie(nullptr);
solve();
return 0;
}
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