洛谷P4555 [国家集训队]最长双回文串(manacher 线段树)

题意

题目链接

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;
}