Submission #1019061

#	Time	Username	Problem	Language	Result	Execution time	Memory
1019061		stefanopulos	Palinilap (COI16_palinilap)	C++17	54 / 100	1065 ms	29532 KiB

palinilap

#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 = 200005; 

int n;
string s;


int d1[mxN];
int d2[mxN];
void manacher(string a){
    int N = sz(a);
    int L = 0, R = -1;
    ff(i,0,N - 1){
        int j = (i > R ? 1 : min(d1[L + R - i], R - i + 1));
        while(i >= j && i + j < N && a[i - j] == a[i + j])j += 1;
        d1[i] = j--;
        if(i + j > R){
            L = i - j;
            R = i + j;
        }
    }
    
    L = 0, R = -1;
    ff(i,0,N - 1){
        int j = (i > R ? 0 : min(d2[L + R - i + 1], R - i + 1));
        while(i > j && i + j < N && a[i - j - 1] == a[i + j])j += 1;
        d2[i] = j--;
        if(i + j > R){
            L = i - j - 1;
            R = i + j;
        }
    }
    
}

int kol[mxN];
vector<int> L1[mxN];
vector<int> R1[mxN];
vector<int> L2[mxN];
vector<int> R2[mxN];

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

    cin >> s; n = sz(s);

    manacher(s);

    int uk = 0;
    ff(i,0,n - 1)uk += d1[i] + d2[i];

    ff(i,0,n - 1){
        L1[i + d1[i] - 1].pb(i);
        L2[i + d2[i] - 1].pb(i); 

        R1[i - d1[i] + 1].pb(i); 
        R2[i - d2[i]].pb(i); 

        kol[i] = uk;
    }

    ff(i,0,n - 1){

        // LEVO            
        ff(j,0,i - 1){
            if(j + d1[j] - 1 >= i){
                kol[i] += i - (j + d1[j]);
            }
        }

        ff(j,0,i){
            if(j + d2[j] - 1 >= i){
                kol[i] += i - (j + d2[j]);
            }
        }


    }

    fb(i,n - 1,0){

        // DESNO
        ff(j,i + 1,n - 1){
            if(j - d1[j] + 1 <= i){
                kol[i] += (j - d1[j]) - i;
            }

        }

        ff(j,i + 1,n - 1){
            if(j - d2[j] <= i){
                kol[i] += (j - 1 - d2[j]) - i;
            }
        }

    }

    ff(i,0,n - 1){

        char old = s[i]; int mx = 0;
        for(char a = 'a'; a <= 'z'; a++){
            if(old == a)continue;
            
            s[i] = a; int cur = 0;

            for(auto j : L1[i - 1]){

                int l = 2 * j - i, r = i;
                while(0 <= l && l < n && 0 <= r && r < n && l <= r && s[l] == s[r]){
                    l -= 1; r += 1;
                    cur += 1;
                }

            }

            for(auto j : L2[i - 1]){

                int l = 2 * j - i - 1, r = i;
                while(0 <= l && l < n && 0 <= r && r < n && l <= r && s[l] == s[r]){
                    l -= 1; r += 1;
                    cur += 1;
                }

            }
            
            for(auto j : R1[i + 1]){
                int l = i, r = 2 * j - i;
                while(0 <= l && l < n && 0 <= r && r < n && l <= r && s[l] == s[r]){
                    l -= 1; r += 1;
                    cur += 1;
                }
            }

            for(auto j : R2[i + 1]){
                int l = i, r = 2 * j - i - 1;
                while(0 <= l && l < n && 0 <= r && r < n && l <= r && s[l] == s[r]){
                    l -= 1; r += 1;
                    cur += 1;
                }
            }

            mx = max(mx, cur);

        }

        s[i] = old; kol[i] += mx;

    }

    int rez = uk;
    ff(i,0,n - 1)rez = max(rez, kol[i]);
    cout << rez << '\n';

    return 0;
}
/*



// probati bojenje sahovski
*/
 
 
 
 
 

• Subtask 1: 17 / 17
• Subtask 2: 37 / 37
• Subtask 3: 0 / 46