Submission #545035

#	Time	Username	Problem	Language	Result	Execution time	Memory
545035		leaked	Sightseeing in Kyoto (JOI22_kyoto)	C++14	40 / 100	1388 ms	260748 KiB

kyoto

#include <bits/stdc++.h>

#define f first
#define s second
#define m_p make_pair
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define vec vector
#define pb push_back
#define sz(x) (int)(x).size()
#define pw(x) (1LL<<(x))
#define fast_resp ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef long double ld;
typedef pair<int,ll> pil;
template<class T> bool umin(T &a,const T &b){return (a>b?a=b,1:0);}
template<class T> bool umax(T &a,const T &b){return (a<b?a=b,1:0);}
const int M=1e9+7;
//const int N=5e5+1;
void add(int &a,int b){
    a+=b;
    if(a>=M) a-=M;
    else if(a<0) a+=M;
}
int mult(int a,int b){
    return 1ll*a*b%M;
}
//int dp[N][26];
const int N=4e3+1;
ll dp[N][N];
signed main(){
    fast_resp;
    int n,m;
    cin>>n>>m;
    vec<int>a(n),b(m);
//    int i=0,j=0;
    for(auto &z : a) cin>>z;
    for(auto &z : b) cin>>z;
    int mn=1e9+1;
    vec<int> p1(n),p2(m);
    vec<int> kx,ky;
    for(int i=0;i<n;i++){
        if(umin(mn,a[i])) kx.pb(i);
    }
    mn=1e9+1;
    for(int i=n-1;i>=0;i--){
        if(umin(mn,a[i])) kx.pb(i);
    }
    mn=1e9+1;
    for(int i=m-1;i>=0;i--){
        if(umin(mn,b[i])) ky.pb(i);
    }
    mn=1e9+1;
    for(int i=0;i<m;i++){
        if(umin(mn,b[i])) ky.pb(i);
    }

    sort(all(kx));kx.erase(unique(all(kx)),kx.end());
    sort(all(ky));ky.erase(unique(all(ky)),ky.end());
//    cout<<endl;
//    cout<<0<<' ';
    for(int i=0;i<sz(kx);i++){
        for(int j=0;j<sz(ky);j++){
            if(!i && !j) continue;
            dp[i][j]=1e18;
            if(i) umin(dp[i][j],dp[i-1][j]+1ll*(kx[i]-kx[i-1])*b[ky[j]]);
            if(j) umin(dp[i][j],dp[i][j-1]+1ll*(ky[j]-ky[j-1])*a[kx[i]]);
//            cout<<dp[i][j]-((i&&j)?dp[i-1][j-1]:0)<<' ';
        }
//        cout<<endl;
    }
    cout<<dp[sz(kx)-1][sz(ky)-1];
    return 0;
}
/*

*/

• Subtask 1: 10 / 10
• Subtask 2: 30 / 30
• Subtask 3: 0 / 60