Submission #1017907

#	Time	Username	Problem	Language	Result	Execution time	Memory
1017907		stefanopulos	Sličnost (COI23_slicnost)	C++17	50 / 100	233 ms	51608 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 ls[200 * mxN], rs[100 * mxN], root[100 * mxN];
int mx[100 * mxN], sum[100 * mxN], cnt[100 * mxN];
void build(int &v, int tl, int tr){
    v = ++idx;
    if(tl == tr){
        mx[v] = sum[v] = 0;
        cnt[v] = 1;
        return;
    }
    int mid = (tl + tr) / 2;
    build(ls[v], tl, mid); build(rs[v], mid + 1, tr);

    sum[v] = sum[ls[v]] + sum[rs[v]];
    mx[v] = max(mx[ls[v]], mx[rs[v]] + sum[ls[v]]);
    cnt[v] = (mx[v] == mx[ls[v]] ? cnt[ls[v]] : 0) + (mx[v] == mx[rs[v]] + sum[ls[v]] ? cnt[rs[v]] : 0);
}

void update(int &v, int rv, int tl, int tr, int pos, int val){
    v = ++idx; ls[v] = ls[rv]; rs[v] = rs[rv]; mx[v] = mx[rv]; sum[v] = sum[rv]; cnt[v] = cnt[rv];
    if(tl == tr){
        mx[v] += val;
        sum[v] += val;
        return;
    }
    int mid = (tl + tr) / 2;
    if(pos <= mid)update(ls[v], ls[rv], tl, mid, pos, val);
    else update(rs[v], rs[rv], mid + 1, tr, pos, val);

    sum[v] = sum[ls[v]] + sum[rs[v]];
    mx[v] = max(mx[ls[v]], mx[rs[v]] + sum[ls[v]]);
    cnt[v] = (mx[v] == mx[ls[v]] ? cnt[ls[v]] : 0) + (mx[v] == mx[rs[v]] + sum[ls[v]] ? cnt[rs[v]] : 0);

}

void update(int &v, int tl, int tr, int pos, int val){
    if(!v){
        v = ++idx;
        mx[v] = sum[v] = 0;
        cnt[v] = 1;
    }

    if(tl == tr){
        mx[v] += val;
        sum[v] += val;
        return;
    }
    int mid = (tl + tr) / 2;
    if(pos <= mid)update(ls[v], tl, mid, pos, val);
    else update(rs[v], mid + 1, tr, pos, val);

    sum[v] = sum[ls[v]] + sum[rs[v]];
    mx[v] = max(mx[ls[v]], mx[rs[v]] + sum[ls[v]]);
    cnt[v] = (mx[v] == mx[ls[v]] ? cnt[ls[v]] : 0) + (mx[v] == mx[rs[v]] + sum[ls[v]] ? cnt[rs[v]] : 0);

}

struct SegTree{
    ll bor[4 * mxN][2];
    void update(int v, int tl, int tr, int pos, int val, ll cnt){
        if(tl == tr){
            bor[v][0] = val;
            bor[v][1] = cnt;
            return;
        }
        int mid = (tl + tr) / 2;
        if(pos <= mid)update(v * 2, tl, mid, pos, val, cnt);
        else update(v * 2 + 1, mid + 1, tr, pos, val, cnt);
        bor[v][0] = max(bor[v * 2][0], bor[v * 2 + 1][0]);
        bor[v][1] = (bor[v][0] == bor[v * 2][0] ? bor[v * 2][1] : 0) + (bor[v][0] == bor[v * 2 + 1][0] ? bor[v * 2 + 1][1] : 0);
    }
}drvo;

void calc(){

    idx = 0;
    build(root[k],1,n - k + 1);
    ff(i,1,k){
        int l = L[poz[A[i]]];
        int r = R[poz[A[i]]];
        update(root[k],root[k],1,n - k + 1,l,1);
        if(r < n - k + 1)update(root[k],1,n - k + 1,r + 1,-1);
    }
    
    drvo.update(1,1,n - k + 1,1,mx[root[k]],cnt[root[k]]);
    ff(i,k + 1,n){
        int l1 = L[poz[A[i - k]]];
        int r1 = R[poz[A[i - k]]];
        update(root[i],root[i - 1],1,n - k + 1,l1,-1);
        if(r1 < n - k + 1)update(root[i],1,n - k + 1,r1 + 1,1);
 
        int l2 = L[poz[A[i]]];
        int r2 = R[poz[A[i]]];
        update(root[i],1,n - k + 1,l2,1);
        if(r2 < n - k + 1)update(root[i],1,n - k + 1,r2 + 1,-1);

        drvo.update(1,1,n - k + 1,i - k + 1,mx[root[i]],cnt[root[i]]);

    }
 
}
 
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();
 
    int najv = drvo.bor[1][0]; ll br = drvo.bor[1][1];
    cout << najv << " " << br << '\n';
    while(q--){
        int t;
        cin >> t;

        if(t - k + 1 >= 1){
            int x = t;
            if(mx[root[x]] == najv)br -= cnt[root[x]];

            int l1 = L[poz[A[t]]];
            int r1 = R[poz[A[t]]];
            update(root[x],1,n - k + 1,l1,-1);
            if(r1 < n - k + 1)update(root[x],1,n - k + 1,r1 + 1,1);

            int l2 = L[poz[A[t + 1]]];
            int r2 = R[poz[A[t + 1]]];
            update(root[x],1,n - k + 1,l2,1);
            if(r2 < n - k + 1)update(root[x],1,n - k + 1,r2 + 1,-1);

            drvo.update(1,1,n - k + 1,x - k + 1,mx[root[x]],cnt[root[x]]);

        }
 
        if(t + k <= n){
            int x = t + k;
            if(mx[root[x]] == najv)br -= cnt[root[x]];

            int l1 = L[poz[A[t + 1]]];
            int r1 = R[poz[A[t + 1]]];
            update(root[x],1,n - k + 1,l1,-1);
            if(r1 < n - k + 1)update(root[x],1,n - k + 1,r1 + 1,1);

            int l2 = L[poz[A[t]]];
            int r2 = R[poz[A[t]]];
            update(root[x],1,n - k + 1,l2,1);
            if(r2 < n - k + 1)update(root[x],1,n - k + 1,r2 + 1,-1);

            drvo.update(1,1,n - k + 1,x - k + 1,mx[root[x]],cnt[root[x]]);

        }

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

        najv = drvo.bor[1][0]; br = drvo.bor[1][1];
        cout << najv << " " << br << '\n';
 
    }
 
    return 0;
}
/*
 
4 3 0
2 4 1 3
1 2 3 4


5 3 1
1 4 3 2 5
4 5 1 2 3
3


 
// probati bojenje sahovski
*/
 
 
 
 
 

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