oj.uz

답안 #503055

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
503055 2022-01-07T04:39:30 Z zaneyu 3D Histogram (COCI20_histogram) C++14
20 / 110
 861 ms 410008 KB 
/*input
6
3 1
2 1
2 2
2 3
1 1
2 2
*/
#include<bits/stdc++.h>
#include <nmmintrin.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<ll,ll>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll maxn=2e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?"\n":"");
    return out;
}
ll mult(ll a,ll b){
    return a*b%MOD;
}
ll mypow(ll a,ll b){
    a%=MOD;
    if(a==0) return 0;
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
int a[maxn],b[maxn];
ll ans=0;
int sa[maxn],sb[maxn];
struct line{
    ll m,c;
    ll eval(ll x){
        return m*x+c;
    }
    ld intersect(line x){
        return (ld)(x.c-c)/(m-x.m);
    }
}; //beware of same slope
struct cht{
    deque<line> dq;
    void add(line z){
        //CHECK THIS
        if(sz(dq) and dq.back().m==z.m) return;
        while(sz(dq)>=2 and dq.back().intersect(dq[sz(dq)-2])>=dq[sz(dq)-2].intersect(z)){
            dq.pop_back();
        }
        dq.pb(z);
    }
    ll query(ll x){
        assert(sz(dq));
        while(sz(dq)>=2 and dq[0].eval(x)<=dq[1].eval(x)) dq.pop_front();
        return dq[0].eval(x);
    }
}seg[3*maxn];
void build(int idx,int l,int r){
    seg[idx].dq.clear();
    for(int i=l;i<=r;i++) seg[idx].add((line){sb[i],-1LL*sb[i]*i});
    if(l==r) return;   
    int mid=(l+r)/2;
    build(idx*2,l,mid),build(idx*2+1,mid+1,r);
}
ll query(int idx,int l,int r,int ql,int qr,int x){
    if(r<ql or l>qr) return 0;
    if(ql<=l and r<=qr) return seg[idx].query(x);
    int mid=(l+r)/2;
    return max(query(idx*2,l,mid,ql,qr,x),query(idx*2+1,mid+1,r,ql,qr,x));
}
void rec(int l,int r){
    if(l==r){
        MXTO(ans,1LL*a[l]*b[l]);
        return;
    }
    int mid=(l+r)/2;
    int ma=INF,mb=INF;
    sa[mid+1]=sb[mid+1]=INF;
    for(int i=mid;i>=l;i--){
        sa[i]=min(sa[i+1],a[i]),sb[i]=min(sb[i+1],b[i]);
    }
    int p=mid;
    for(int i=mid+1;i<=r;i++){
        MNTO(ma,a[i]),MNTO(mb,b[i]);
        while(p>=l and (sa[p]>=ma and sb[p]>=mb)) --p;
        MXTO(ans,1LL*(i-p)*ma*mb);
    }
    ma=INF,mb=INF;
    //min a on right
    //sa[l]>pa[r] sb[l]<pb[r]
    //forms a left bound, forms a right bound
    //left bound decreases, right bound decreases 
    //pain
    int x=mid,y=mid;
    build(1,l,mid);
    for(int i=mid+1;i<=r;i++){
        MNTO(ma,a[i]),MNTO(mb,b[i]);
        while(x>=l and sa[x]>ma) --x;
        while(y>=l and sb[y]>mb) --y;
        if(x+1<=y) MXTO(ans,1LL*query(1,l,mid,x+1,y,i+1)*ma);
    }
    rec(l,mid),rec(mid+1,r);
}
int32_t main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n;
    cin>>n;
    REP(i,n){
        cin>>a[i]>>b[i];
    }
    rec(0,n-1);
    reverse(a,a+n),reverse(b,b+n);
    rec(0,n-1);
    cout<<ans;
}

Subtask #1
20.0 / 20.0

# 결과 실행 시간 메모리 Grader output
1 Correct 233 ms 404248 KB Output is correct
2 Correct 226 ms 404292 KB Output is correct
3 Correct 233 ms 404168 KB Output is correct
4 Correct 252 ms 404192 KB Output is correct
5 Correct 226 ms 404256 KB Output is correct
6 Correct 239 ms 404292 KB Output is correct
7 Correct 232 ms 404164 KB Output is correct
8 Correct 226 ms 404224 KB Output is correct
9 Correct 231 ms 404284 KB Output is correct
10 Correct 246 ms 404196 KB Output is correct
11 Correct 226 ms 404204 KB Output is correct
12 Correct 253 ms 404296 KB Output is correct

Subtask #2
0 / 90.0

# 결과 실행 시간 메모리 Grader output
1 Correct 233 ms 404248 KB Output is correct
2 Correct 226 ms 404292 KB Output is correct
3 Correct 233 ms 404168 KB Output is correct
4 Correct 252 ms 404192 KB Output is correct
5 Correct 226 ms 404256 KB Output is correct
6 Correct 239 ms 404292 KB Output is correct
7 Correct 232 ms 404164 KB Output is correct
8 Correct 226 ms 404224 KB Output is correct
9 Correct 231 ms 404284 KB Output is correct
10 Correct 246 ms 404196 KB Output is correct
11 Correct 226 ms 404204 KB Output is correct
12 Correct 253 ms 404296 KB Output is correct
13 Correct 700 ms 409228 KB Output is correct
14 Correct 695 ms 409932 KB Output is correct
15 Correct 610 ms 408816 KB Output is correct
16 Correct 695 ms 408836 KB Output is correct
17 Correct 732 ms 409936 KB Output is correct
18 Correct 861 ms 410000 KB Output is correct
19 Correct 792 ms 409912 KB Output is correct
20 Correct 788 ms 410008 KB Output is correct
21 Correct 661 ms 408832 KB Output is correct
22 Correct 769 ms 409940 KB Output is correct
23 Incorrect 255 ms 404548 KB Output isn't correct
24 Halted 0 ms 0 KB -