题意
Sol
我的做法比较naive。。首先manacher预处理出以每个位置为中心的回文串的长度。然后枚举一个中间位置，现在要考虑的就是能覆盖到i - 1的回文串中 中心最靠左的，和能覆盖到i+1中 中心最靠右的，算一下答案取个max。
线段树维护一下区间min, max。标记永久化炒鸡好写
// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
char s[MAXN];
int len[MAXN], N, ans[MAXN];
template<typename A, typename B> inline void chmax(A &x, B y) {
x = x < y ? y : x;
}
template<typename A, typename B> inline void chmin(A &x, B y) {
x = x < y ? x : y;
}
int root, ls[MAXN], rs[MAXN], mn[MAXN], mx[MAXN], tot;
void Max(int &k, int l, int r, int ql, int qr, int v) {
if(!k) k = ++tot, mn[k] = INF;
if(ql <= l && r <= qr) {chmax(mx[k], v); return ;}
int mid = l + r >> 1;
if(ql <= mid) Max(ls[k], l, mid, ql, qr, v);
if(qr > mid) Max(rs[k], mid + 1, r, ql, qr, v);
}
void Min(int &k, int l, int r, int ql, int qr, int v) {
if(!k) k = ++tot, mn[k] = INF;
if(ql <= l && r <= qr) {chmin(mn[k], v); return ;}
int mid = l + r >> 1;
if(ql <= mid) Min(ls[k], l, mid, ql, qr, v);
if(qr > mid) Min(rs[k], mid + 1, r, ql, qr, v);
}
int QueryMx(int k, int l, int r, int p) {
int ans = mx[k];
if(l == r) return ans;
int mid = l + r >> 1;
if(p <= mid) chmax(ans, QueryMx(ls[k], l, mid, p));
else chmax(ans, QueryMx(rs[k], mid + 1, r, p));
return ans;
}
int QueryMn(int k, int l, int r, int p) {
int ans = mn[k];
if(l == r) return ans;
int mid = l + r >> 1;
if(p <= mid) chmin(ans, QueryMn(ls[k], l, mid, p));
else chmin(ans, QueryMn(rs[k], mid + 1, r, p));
return ans;
}
void trans() {
static char tmp[MAXN];
for(int i = 1; i <= N; i++) {
tmp[2 * i - 1] = s[i];
tmp[2 * i] = '#';
}
memcpy(s, tmp, sizeof(s));
N = (N << 1) - 1;
int mx = 0, id = 0;
for(int i = 1; i <= N; i++) {
ans[i] = (mx > i ? min(mx - i, ans[id * 2 - i]) : 1);
while(s[i - ans[i]] == s[i + ans[i]]) ans[i]++;
if(i + ans[i] > mx) mx = i + ans[i], id = i;
Max(root, 1, N, i - ans[i] + 1, i, i);
Min(root, 1, N, i, i + ans[i] - 1, i);
}
}
int main() {
scanf("%s", s + 1);
N = strlen(s + 1);
trans();
int ans = 0;
for(int i = 2; i <= N; i += 2) {
chmax(ans, (i - 1 - QueryMn(root, 1, N, i - 1)) + 1 + (QueryMx(root, 1, N, i + 1) - i - 1) + 1);
}
cout << ans;
return 0;
}