Submission #567415

#	Time	Username	Problem	Language	Result	Execution time	Memory
567415		zaneyu	Sightseeing in Kyoto (JOI22_kyoto)	C++14	100 / 100	425 ms	13724 KiB

kyoto

/*input
4 6
454863204 543362989 866044086 813602010
71574269 17945210 688720933 392135202 38174709 168241720
*/
#include<bits/stdc++.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<int,int>;
#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(int 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=5e5+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)?" ":"");
    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];
#define iii pair<double,pii> 
int main(){
    int n,m;
    cin>>n>>m;
    REP(i,n) cin>>a[i];
    REP(i,m) cin>>b[i];
    priority_queue<iii,vector<iii>,greater<iii>> pq,pq2;
    REP1(i,n-1) pq.push({a[i-1]-a[i],{i-1,i}});
    REP1(i,m-1) pq2.push({b[i-1]-b[i],{i-1,i}});
    set<int> s1,s2;
    REP(i,n) s1.insert(i);
    REP(i,m) s2.insert(i);
    ll ans=0;
    while(sz(pq)+sz(pq2)){
        if(sz(pq) and (!sz(pq2) or pq.top()<pq2.top())){
            pii aa=pq.top().s;
            pq.pop();
            if(!s1.count(aa.f) or !s1.count(aa.s)) continue;
            int l=aa.f,r=-1;
            auto it=s1.find(aa.s);
            if(it!=prev(s1.end())) r=*next(it);
            s1.erase(it);
            if(r!=-1) pq.push({(ld)(a[l]-a[r])/(r-l),{l,r}});
            else ans+=1LL*b[(*s2.rbegin())]*(aa.s-aa.f);
        }
        else{
            pii aa=pq2.top().s;
            pq2.pop();
            if(!s2.count(aa.f) or !s2.count(aa.s)) continue;
            int l=aa.f,r=-1;
            auto it=s2.find(aa.s);
            if(it!=prev(s2.end())) r=*next(it);
            s2.erase(it);
            if(r!=-1) pq2.push({(ld)(b[l]-b[r])/(r-l),{l,r}});
            else ans+=1LL*a[(*s1.rbegin())]*(aa.s-aa.f);   
        }
    }
    cout<<ans;
} 

• Subtask 1: 10 / 10
• Subtask 2: 30 / 30
• Subtask 3: 60 / 60