oj.uz

답안 #1019234

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
1019234 2024-07-10T16:03:06 Z stefanopulos Palinilap (COI16_palinilap) C++17
100 / 100
 104 ms 44892 KB 
#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;
        }
    }
    
}

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

vector<int> in[mxN];

int add(int a, int b){ a += b; if(a >= mod)a -= mod; return a; }
int sub(int a, int b){ a -= b; if(a < 0)a += mod; return a; }
int mul(int a, int b){ return (1ll * a * b) % mod; }
int power(int a, int b){
    if(!b)return 1;
    int p = power(a, b / 2);
    p = mul(p, p);
    if(b % 2 == 1)p = mul(p, a);
    return p;
}
int inv(int a){
    return power(a, mod - 2);
}

const int p1 = 37;
int pw[mxN];
int invz[mxN];

int pref[mxN];
int getP(int l, int r){
    return mul(sub(pref[r], (l == 0 ? 0 : pref[l - 1])), invz[l]); 
}

int sufi[mxN];
int getS(int l, int r){
    return mul(sub(sufi[l], sufi[r + 1]), invz[n - r - 1]); 
}

int calc(int a, int b ){

    int l = 1, r = n, ans = 0;
    while(l <= r){
        int mid = (l + r) / 2;
        if(a - mid + 1 >= 0 && b + mid - 1 < n && getP(a - mid + 1, a) == getS(b, b + mid - 1)){
            ans = mid;
            l = mid + 1;
        }
        else r = mid - 1;
    }

    return ans;

}

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

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

    pw[0] = 1;
    ff(i,1,n)pw[i] = mul(pw[i - 1], p1);

    invz[n] = inv(pw[n]);
    fb(i,n - 1,0)invz[i] = mul(invz[i + 1], p1);

    ff(i,0,n - 1)pref[i] = add((i == 0 ? 0 : pref[i - 1]), mul(s[i] - 'a' + 1, pw[i]));
    fb(i,n - 1,0)sufi[i] = add(sufi[i + 1], mul(s[i] - 'a' + 1, pw[n - i - 1]));

    manacher(s);

    ll 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;
    }


    ll sum = 0, br = 0;
    ff(i,0,n - 1){
        if(d2[i] > 0){
            in[i + d2[i] - 1].pb(i + d2[i]); sum += (i + d2[i]); br += 1;
        }
        
        kol[i] += 1ll * br * i - sum;

        in[i + d1[i] - 1].pb(i + d1[i]); sum += (i + d1[i]); br += 1;
        for(auto c : in[i]){
            sum -= c; br -= 1;
        }

        in[i].clear();

    }

    sum = 0, br = 0;
    fb(i,n - 1,0){
        kol[i] += sum - 1ll * br * i;

        in[i - d1[i] + 1].pb(i - d1[i]); sum += (i - d1[i]); br += 1;
        if(d2[i] > 0){
            in[i - d2[i]].pb(i - 1 - d2[i]); sum += (i - 1 - d2[i]); br += 1;
        }

        for(auto c : in[i]){
            sum -= c; br -= 1;
        }

    }
    
    ff(i,0,n - 1){

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

            for(auto j : L1[i - 1]){
                int l = 2 * j - i, r = i;
                if(0 <= l && l < n && 0 <= r && r < n && l <= r && s[l] == s[r]){
                    l -= 1; r += 1;
                    cur += 1 + calc(l, r);
                    
                }
            }

            for(auto j : L2[i - 1]){
                int l = 2 * j - i - 1, r = i;
                if(0 <= l && l < n && 0 <= r && r < n && l <= r && s[l] == s[r]){
                    l -= 1; r += 1;
                    cur += 1 + calc(l, r);
                }
            }
            
            for(auto j : R1[i + 1]){
                int l = i, r = 2 * j - i;
                if(0 <= l && l < n && 0 <= r && r < n && l <= r && s[l] == s[r]){
                    l -= 1; r += 1;
                    cur += 1 + calc(l, r);
                }
            }

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

            mx = max(mx, cur);

        }

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

    }

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

    return 0;
}
/*



// probati bojenje sahovski
*/

Subtask #1
17.0 / 17.0

# 결과 실행 시간 메모리 Grader output
1 Correct 4 ms 29020 KB Output is correct
2 Correct 5 ms 29020 KB Output is correct
3 Correct 4 ms 29212 KB Output is correct
4 Correct 6 ms 29020 KB Output is correct
5 Correct 5 ms 29016 KB Output is correct

Subtask #2
37.0 / 37.0

# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 29784 KB Output is correct
2 Correct 7 ms 29784 KB Output is correct
3 Correct 8 ms 29788 KB Output is correct
4 Correct 6 ms 29532 KB Output is correct
5 Correct 8 ms 29788 KB Output is correct
6 Correct 9 ms 29768 KB Output is correct
7 Correct 8 ms 29784 KB Output is correct

Subtask #3
46.0 / 46.0

# 결과 실행 시간 메모리 Grader output
1 Correct 99 ms 42076 KB Output is correct
2 Correct 60 ms 41172 KB Output is correct
3 Correct 53 ms 41296 KB Output is correct
4 Correct 96 ms 42304 KB Output is correct
5 Correct 97 ms 42568 KB Output is correct
6 Correct 97 ms 42512 KB Output is correct
7 Correct 101 ms 42504 KB Output is correct
8 Correct 43 ms 41160 KB Output is correct
9 Correct 104 ms 42332 KB Output is correct
10 Correct 101 ms 42320 KB Output is correct
11 Correct 80 ms 41172 KB Output is correct
12 Correct 80 ms 41820 KB Output is correct
13 Correct 86 ms 41960 KB Output is correct
14 Correct 90 ms 42832 KB Output is correct
15 Correct 93 ms 42584 KB Output is correct
16 Correct 85 ms 41036 KB Output is correct
17 Correct 90 ms 44892 KB Output is correct
18 Correct 94 ms 42844 KB Output is correct
19 Correct 90 ms 44848 KB Output is correct