Submission #1266640

#	Time	Username	Problem	Language	Result	Execution time	Memory
1266640		syanvu	Group Photo (JOI21_ho_t3)	C++20	0 / 100	0 ms	324 KiB

Main

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp>

#define SS ios_base::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);
#pragma optimize("g", on)
#pragma GCC optimize("03")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,avx2,mmx,fma,avx,tune=native")
#define int long long
#define all(x) x.begin(),x.end()
#define F first
#define S second


using namespace std;
// using namespace __gnu_pbds; 
// #define ordered_set tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>

const int LG = 21,N = 2e5+1,inf = 1e18,MOD = 998244353;
const double eps = 1e-9;
int T;

int f[N];
int n;

void add(int i,int x){
    while(i<=n){
        f[i]+=x;
        i+=i&-i;
    }
}
int get(int i){
    int res=0;
    while(i){
        res+=f[i];
        i-=i&-i;
    }
    return res;
}
int range(int l,int r){
    return get(r)-get(l);
}

void solve() {
    cin>>n;
    int h[n+1];
    int id[n+1];
    for(int i=1;i<=n;i++){
        cin>>h[i];
        id[h[i]]=i;
    }
    int cost[n+2][n+2]={},pref[n+2][n+2]={};
    for(int r=1;r<=n;r++){
        for(int i=1;i<=r;i++) pref[r][id[i]]++;
        for(int i=1;i<=n;i++){
            pref[r][i]+=pref[r][i-1];
        }
        for(int l=r;l>=1;l--){
            cost[l][r]=cost[l+1][r]+(pref[r][n]-pref[r][id[l]])-get(id[l]);
            if(r<id[l]) cost[l][r]+=id[l]-r;
            add(id[l],1);
        }
        for(int i=1;i<=n;i++) f[i]=0;
    }
    vector<int> dp(n+1,inf);
    dp[0]=0;
    for(int r=1;r<=n;r++){
        for(int l=r;l>=1;l--){
            dp[r]=min(dp[r],dp[l-1]+cost[l][r]);
        }
    }
    cout<<dp[n];
    return;
}
signed main() {
    //   freopen("deleg.in","r",stdin);    
    // freopen("deleg.out","w",stdout);
    SS
    int t = 1;
    if (T) {
        cin >> t;
    }
    while (t--) {
        solve();
    }
}
/*
1,2,3,...,n-2,n-1,n

good array is reverse of subsequence of subarray
because the condition a[i]-a[i+1]<=1 must hold
    PROOF
    {
        Suppose on the position i there is number x
        the next number which we can pick is x-1 or y>x
        when we take number which is greater than x, x-1 number must be taken on prefix
        Otherwise y>x will only decrease by 1 and the minimum number on suffix will x+1, and x+1-x-1=2 condition doesnt hold
    }
dp[i][j] -> prefix i, last number on reversed array

1<=j<=i dp[i] = dp[j-1] + op[j]


*/

• Subtask 1: 0 / 5
• Subtask 2: 0 / 7
• Subtask 3: 0 / 32
• Subtask 4: 0 / 20
• Subtask 5: 0 / 36