#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define sz(a) (int)(a.size())
#define all(a) a.begin(),a.end()
#define lb lower_bound
#define ub upper_bound
#define owo ios_base::sync_with_stdio(0);cin.tie(0);
#define MOD (ll)(998244353)
#define INF (ll)(1e18)
#define debug(...) fprintf(stderr, __VA_ARGS__),fflush(stderr)
#define time__(d) for(long blockTime = 0; (blockTime == 0 ? (blockTime=clock()) != 0 : false);\
debug("%s time : %.4fs\n", d, (double)(clock() - blockTime) / CLOCKS_PER_SEC))
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> PII;
typedef pair<int,int> pii;
typedef vector<vector<int>> vii;
typedef vector<vector<ll>> VII;
ll gcd(ll a,ll b){if(!b)return a;else return gcd(b,a%b);}
const int MAXN = 2000001;
int a[MAXN],b[MAXN],f[MAXN],g[MAXN],n;
ll ans = 0;
ll d(ll a, ll b){return a / b + ((a^b) < 0 && (a%b));}
struct line{
ll m,c;
ll eval(ll x){return m*x+c;}
line(ll _m,ll _c):m(_m),c(_c){}
line(){}
};
struct hull{
vector<line>h;
void add(line L){
while(sz(h) >= 2){
line c0 = h[sz(h)-1];
line c1 = h[sz(h)-2];
if(d(c0.c - c1.c , c1.m - c0.m) >= d(c0.c - L.c , L.m - c0.m))h.pop_back();
else break;
}
h.pb(L);
}
ll query(ll x){
while(sz(h)>=2 && h[sz(h)-2].eval(x) >= h[sz(h)-1].eval(x))h.pop_back();
return h.back().eval(x);
}
};
hull t[4*MAXN];
void build(int v,int l,int r,int tl,int tr){
if(l>r)return;
if(l==tl && r==tr){
t[v].h.clear();
for(int i=r;i>=l;i--){
if(i!=r && g[i] == g[i+1])continue;
t[v].add(line(g[i],g[i] * 1LL * i));
}
return;
}else{
int tm = (tl+tr)/2;
build(2*v,l,min(r,tm),tl,tm);
build(2*v+1,max(tm+1,l),r,tm+1,tr);
}
}
ll query(int v,int l,int r,int tl,int tr,ll x){
if(l>r)return 0;
if(l==tl && r==tr)return t[v].query(x);
else{
int tm = (tl+tr)/2;
return max(query(2*v,l,min(r,tm),tl,tm,x),
query(2*v+1,max(tm+1,l),r,tm+1,tr,x));
}
}
void solve(int l,int r){
if(l==r)return;
int m = (l+r)/2;
solve(l,m);
solve(m+1,r);
f[m] = a[m];f[m+1] = a[m+1];
g[m] = b[m];g[m+1] = b[m+1];
for(int i=m-1;i>=l;i--)f[i] = min(a[i],f[i+1]);
for(int i=m+2;i<=r;i++)f[i] = min(a[i],f[i-1]);
for(int i=m+2;i<=r;i++)g[i] = min(b[i],g[i-1]);
for(int i=m-1;i>=l;i--)g[i] = min(b[i],g[i+1]);
int p = r;
for(int i=l;i<=m;i++){ //f and g lie on left side
while(p>m && (f[i] > f[p] || g[i] > g[p]))p--;
if(p==m)break;
ans = max(ans,f[i] * 1LL * g[i] * 1LL * (p-i+1));
}
//f lie on left and g left on right
//f[i] * g[r] * (r-l+1)
//f[i] (g[r]*(-l+1) + g[r]*r)
//g is monotonically decreasing we can just insert backwards
build(1,m+1,r,1,n);
int pf = r;
int pg = m+1;
for(int i=l;i<=m;i++){
while(pf>m && f[i] > f[pf])pf--; //[m+1,pf] is good
while(pg<=r && g[i] < g[pg])pg++; //[pg,r] is good
if(pf == m || pg > r)break;
ans = max(ans,f[i] * query(1,pg,pf,1,n,-i+1));
}
}
int main(){
owo
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i]>>b[i];
ans = max(ans,a[i] * 1LL * b[i]);
}
solve(1,n);
reverse(a+1,a+n+1);
reverse(b+1,b+n+1);
solve(1,n);
cout<<ans;
}
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Runtime error |
246 ms |
381380 KB |
Execution killed with signal 11 |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Runtime error |
246 ms |
381380 KB |
Execution killed with signal 11 |
| 2 |
Halted |
0 ms |
0 KB |
- |
