Submission #978800

#	Time	Username	Problem	Language	Result	Execution time	Memory
978800		stefanopulos	Sličnost (COI23_slicnost)	C++17	17 / 100	375 ms	278644 KiB

Main

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ldb;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ldb,ldb> pdd;

#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for(auto& a : x)
 
#define sz(a) (int)(a).size()
#define fi first
#define se second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 100005; 

int n, k, q;
int A[mxN];
int B[mxN];

int poz[mxN];

int L[mxN];
int R[mxN];

int idx = 0;
int root[100 * mxN], ls[100 * mxN], rs[100 * mxN], lenj[100 * mxN]; ll bor[100 * mxN][2];

void build(int& v, int tl, int tr){
    v = ++idx;

    lenj[v] = 0;
    if(tl == tr){
        bor[v][0] = 0;
        bor[v][1] = 1;
        return;
    }

    int mid = (tl + tr) / 2;
    build(ls[v], tl, mid); build(rs[v], mid + 1, tr);
    bor[v][0] = max(bor[ls[v]][0], bor[rs[v]][0]);
    bor[v][1] = (bor[ls[v]][0] == bor[v][0] ? bor[ls[v]][1] : 0) +
                (bor[rs[v]][0] == bor[v][0] ? bor[rs[v]][1] : 0);
}

void propagate(int v, int tl, int tr){
    if(lenj[v] != 0){
        bor[v][0] += lenj[v];
        if(tl != tr){
            lenj[ls[v]] += lenj[v];
            lenj[rs[v]] += lenj[v];
        }
        lenj[v] = 0;
    }
}

void lazyupd(int& v, int rv, int tl, int tr, int l, int r, int val){
    v = ++idx; ls[v] = ls[rv]; rs[v] = rs[rv]; 
    bor[v][0] = bor[rv][0]; bor[v][1] = bor[rv][1]; lenj[v] += lenj[rv];

    propagate(v, tl, tr);
    if(tl > tr || tl > r || l > tr)return;
    if(tl >= l && tr <= r){
        lenj[v] += val;
        propagate(v, tl, tr);
        return;
    }

    int mid = (tl + tr) / 2;
    lazyupd(ls[v], ls[rv], tl, mid, l, r, val); lazyupd(rs[v], rs[rv], mid + 1, tr, l, r, val);
    bor[v][0] = max(bor[ls[v]][0], bor[rs[v]][0]);
    bor[v][1] = (bor[ls[v]][0] == bor[v][0] ? bor[ls[v]][1] : 0) +
                (bor[rs[v]][0] == bor[v][0] ? bor[rs[v]][1] : 0);
}

void calc(){
    idx = 0;

    build(root[1],1,n - k + 1);
    ff(i,1,k){
        int l = L[poz[A[i]]];
        int r = R[poz[A[i]]];
        lazyupd(root[1],root[1],1,n - k + 1,l,r,1);
    }
    
    int mx = bor[root[1]][0]; ll br = bor[root[1]][1];
    ff(i,k + 1,n){
        int l1 = L[poz[A[i - k]]];
        int r1 = R[poz[A[i - k]]];
        lazyupd(root[i - k + 1],root[i - k],1,n - k + 1,l1,r1,-1);

        int l2 = L[poz[A[i]]];
        int r2 = R[poz[A[i]]];
        lazyupd(root[i - k + 1],root[i - k + 1],1,n - k + 1,l2,r2,1);

        if(bor[root[i - k + 1]][0] > mx){
            mx = bor[root[i - k + 1]][0];
            br = bor[root[i - k + 1]][1];
        }
        else if(bor[root[i - k + 1]][0] == mx)br += bor[root[i - k + 1]][1];

    }

    cout << mx << " " << br << '\n';

}
 
int main(){
    cin.tie(0)->sync_with_stdio(0);

    cin >> n >> k >> q;
    ff(i,1,n)cin >> A[i];
    ff(i,1,n)cin >> B[i], poz[B[i]] = i;

    ff(i,1,n){
        L[i] = max(1, i - k + 1);
        R[i] = min(i, n - k + 1);
    }

    calc();

    while(q--){
        int t;
        cin >> t;

        // int a = A[t];
        // int b = A[t + 1];

        // int x = t - k + 1;
        // if(x >= 1){

        // }


        // int y = t + k - 1;

        swap(A[t], A[t + 1]);
        calc();

    }

    return 0;
}
/*

4 3 0
2 4 1 3
1 2 3 4

// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 7 / 7
• Subtask 2: 10 / 10
• Subtask 3: 0 / 33
• Subtask 4: 0 / 7
• Subtask 5: 0 / 10
• Subtask 6: 0 / 33