题目描述
https://codeforces.com/problemset/problem/961/F
简要题意：定义 $S$ 的 $k$ 子串为 $S_k=S[k..n-k+1],k\le \lceil\frac{n}{2}\rceil$，求 $S_k$ 的最长 $border$ 长度，$k\in [1,\lceil\frac{n}{2}\rceil]$
$|S|\le 10^6$
Solution
令 $ans_i$ 表示 $S_i$ 的最长 $border$，容易发现 $ans_i\le ans_{i+1}+2$，画图容易验证，那么我们暴力求 $ans_i$ 即可，判断相等使用 $hash$ 即可
时间复杂度 $O(n)$
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| #include <iostream>
#include <cstring>
#define maxn 1000010
#define ll long long
using namespace std;
struct Hash {
const int base = 131;
const ll p = 212370440130137957;
ll pow[maxn], h[maxn];
inline ll mul(ll x, ll y) {
return (x * y - (ll) ((long double) x / p * y) * p + p) % p;
}
inline ll add(ll x, ll y) { return (x += y) >= p ? x - p : x; }
void init(char *s) {
int l = strlen(s + 1);
pow[0] = 1; for (int i = 1; i <= l; ++i) pow[i] = mul(pow[i - 1], base);
for (int i = 1; i <= l; ++i)
h[i] = add(s[i], mul(h[i - 1], base));
}
ll get(int l, int r) { return add(h[r], p - mul(h[l - 1], pow[r - l + 1])); }
ll hash(char *s) {
int l = strlen(s + 1); ll ans = 0;
for (int i = 1; i <= l; ++i)
ans = add(s[i], mul(ans, base));
return ans;
}
} H;
int n;
char s[maxn];
int ans[maxn];
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cin >> n >> s + 1; H.init(s);
int m = (n + 1) / 2; ans[m] = (n & 1) ? 0 : s[n / 2] == s[n / 2 + 1];
for (int i = m - 1; i; --i) {
ans[i] = ans[i + 1] + 2;
while (i + ans[i] - 1 > n - i + 1) --ans[i];
while (ans[i] > 0) {
if (i + ans[i] - 1 > n - i + 1 || (~ans[i] & 1) ||
H.get(i, i + ans[i] - 1) != H.get(n - i + 1 - ans[i] + 1, n - i + 1)) --ans[i];
else break;
}
}
for (int i = 1; i <= m; ++i) {
if (!ans[i]) ans[i] = -1;
cout << ans[i] << " \n"[i == m];
}
return 0;
}
|