제출 #388114

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
388114		Keshi	회문 (APIO14_palindrome)	C++17	73 / 100	1094 ms	57908 KiB

palindrome

//In the name of God
#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;

typedef int ll;
typedef pair<ll, ll> pll;

const ll maxn = 3e5 + 100;
const ll lg = 19;
const ll mod = 1e9 + 7;
const ll inf = 1e9;

#define fast_io ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io freopen("input.txt", "r+", stdin);freopen("output.txt", "w+", stdout);
#define pb push_back
#define Mp make_pair
#define F first
#define S second
#define Sz(x) ll((x).size())
#define all(x) (x).begin(), (x).end()
#define lc (id << 1)
#define rc (lc | 1)

ll n, p[maxn], rnk[maxn], r2[maxn], lcp[maxn], f[maxn], per[maxn], ps[maxn], p2[maxn];
string s;
vector<ll> vec[maxn];

ll mn[lg][maxn];

void bld(){
	for(ll i = 0; i < n; i++){
		mn[0][i] = lcp[i];
	}
	for(ll i = 1; i < lg; i++){
		for(ll j = 0; j + (1 << i) <= n; j++){
			mn[i][j] = min(mn[i - 1][j], mn[i - 1][j + (1 << (i - 1))]);
		}
	}
}
ll get(ll l, ll r){
	if(l == r) return inf;
	ll t = r - l;
	ll i = f[t];
	return min(mn[i][l], mn[i][r - (1 << i)]);
}

int main(){
	fast_io;
	for(ll i = 2; i < maxn; i++){
		f[i] = f[i / 2] + 1;
	}

	cin >> s;
	s += '#';
	n = Sz(s);
    ll cnt = 300;
    for(ll i = 0; i < n; i++){
        ps[(ll)s[i]]++;
    }
    for(ll i = 1; i < cnt; i++){
        ps[i] += ps[i - 1];
    }
    for(ll i = n; i--;){
        per[--ps[(ll)s[i]]] = i;
    }
    rnk[per[0]] = 0;
    cnt = 1;
    for(ll i = 1; i < n; i++){
        if(s[per[i - 1]] != s[per[i]]) cnt++;
        rnk[per[i]] = cnt - 1;
    }
    for(ll h = 0; (1 << h) < n; h++){
        fill(ps, ps + cnt, 0);
        for(ll i = 0; i < n; i++){
            p2[i] = per[i] - (1 << h);
            if(p2[i] < 0) p2[i] += n;
        }
        for(ll i = 0; i < n; i++){
            ps[rnk[p2[i]]]++;
        }
        for(ll i = 1; i < cnt; i++){
            ps[i] += ps[i - 1];
        }
        for(ll i = n; i--;){
            per[--ps[rnk[p2[i]]]] = p2[i];
        }
        for(ll i = 0; i < n; i++){
            p2[i] = per[i] + (1 << h);
            if(p2[i] >= n) p2[i] -= n;
        }
        r2[per[0]] = 0;
        cnt = 1;
        for(ll i = 1; i < n; i++){
            if(rnk[per[i]] != rnk[per[i - 1]] || rnk[p2[i]] != rnk[p2[i - 1]]) cnt++;
            r2[per[i]] = cnt - 1;
        }
        for(ll i = 0; i < n; i++){
            rnk[i] = r2[i];
        }
    }
	for(ll i = 0; i < n; i++){
		p[i] = per[i];
	}
	for(ll i = 0, k = 0; i < n; i++){
		if(rnk[i] == n - 1){
			k = 0;
			continue;
		}
		ll j = p[rnk[i] + 1];
		while(i + k < n && j + k < n && s[i + k] == s[j + k]) k++;
		lcp[rnk[i]] = k;
		if(k) k--;
	}
	bld();
	n--;
	vector<int> d1(n);
	for (int i = 0, l = 0, r = -1; i < n; i++) {
   	 	int k = (i > r) ? 1 : min(d1[l + r - i], r - i + 1);
   	 	while (0 <= i - k && i + k < n && s[i - k] == s[i + k]) {
        k++;
    	}
    	d1[i] = k--;
    	if (i + k > r) {
       	 	l = i - k;
        	r = i + k;
    	}
	}
	vector<int> d2(n);
	for (int i = 0, l = 0, r = -1; i < n; i++) {
   		int k = (i > r) ? 0 : min(d2[l + r - i + 1], r - i + 1);
    	while (0 <= i - k - 1 && i + k < n && s[i - k - 1] == s[i + k]) {
        	k++;
    	}
    	d2[i] = k--;
   		if (i + k > r) {
        	l = i - k - 1;
        	r = i + k;
    	}
	}
	set<ll> st;
	long long ans = 0;
	for(ll i = n; i--;){
		for(ll j : vec[i]){
			st.erase(j);
		}
		vec[i].clear();
		st.insert(i);
		vec[i - d1[i]].pb(i);
		auto it = st.rbegin();
		while(it != st.rend()){
		ll t = (*it - i) * 2 + 1;
		if(t <= lcp[rnk[i] - 1]) break;;
		ll l = rnk[i], r = n + 1;
		while(r - l > 1){
			ll mid = (l + r) >> 1;
			if(get(rnk[i], mid) >= t) l = mid;
			else r = mid;
		}
		ans = max(ans, ((long long)t) * (r - rnk[i]));
		it++;
		}
	}
	st.clear();
	for(ll i = n - 1; i--;){
		for(ll j : vec[i]){
			st.erase(j);
		}
		if(d2[i + 1]) st.insert(i);
		if(d2[i + 1]) vec[i - d2[i + 1]].pb(i);
		if(st.empty()) continue;
		auto it = st.rbegin();
		while(it != st.rend()){
		ll t = (*it - i) * 2 + 2;
		if(t <= lcp[rnk[i] - 1]) break;
		ll l = rnk[i], r = n + 1;
		while(r - l > 1){
			ll mid = (l + r) >> 1;
			if(get(rnk[i], mid) >= t) l = mid;
			else r = mid;
		}
		ans = max(ans, ((long long)t) * (r - rnk[i]));
		it++;
		}
	}
	cout << ans;

    return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 15 / 15
• Subtask 3: 24 / 24
• Subtask 4: 26 / 26
• Subtask 5: 0 / 27