【题解】APIO2014回文串

　　哇哦~想不到我有生之年竟然能够做出字符串的题目ヾ(✿ﾟ▽ﾟ)ノ虽然这题比较裸但依然灰常开心！

　　首先有一个棒棒的性质：本质不同的回文串最多有 O(n) 个。首先 manacher 把它们都找出来，然后问题就变成了给定 n 个子串，求它们在原串中出现的次数。求出 height 然后二分一下即可（这个好像是SA 的基础操作？）。

#include <bits/stdc++.h>
using namespace std;
#define maxn 601550
#define CNST 22
int n, N, m, p1[maxn], p2[maxn], rk[maxn], bits[CNST];
int p[maxn], rec[maxn], t[maxn], y[maxn], Log[maxn];
int num[maxn], SA[maxn], ST[maxn][CNST];
long long ans;
char a[maxn], s[maxn];

void pre()
{
    s[] = s[] = '#'; s[ * n + ] = '?';
    for(int i = ; i < n; i ++) s[i *  + ] = a[i], s[i *  + ] = '#';
    N = n *  + ;
}

void Up(int &x, int y) { x = (x > y) ? x : y; }
void manacher()
{
    int mid = , mr = ;
    for(int i = ; i < N; i ++)
    {
        if(i <= mr) p[i] = min(p[(mid << ) - i], p[mid] + mid - i);
        else p[i] = ; Up(rec[i + p[i] - ], p[i]);
        while(s[i + p[i]] == s[i - p[i]])
        {
            p[i] ++;
            Up(rec[i + p[i] - ], p[i]);
        }
        if(p[i] + i -  > mr) mr = p[i] + i - , mid = i;
    }
}

//SA
void Rsort(int *p, int *x, int *id)
{
    for(int i = ; i <= m; i ++) t[i] = ;
    for(int i = ; i <= n; i ++) t[p[i]] ++;
    for(int i = ; i <= m; i ++) t[i] += t[i - ];
    for(int i = n; i >= ; i --) x[t[p[id[i]]] --] = id[i];
}

bool cmp(int x, int y) { return y && (p1[x] == p1[y] && p2[x] == p2[y]); }
void Get_SA()
{
    m = ;
    for(int k = ; bits[k] <= n; k ++)
    {
        for(int i = ; i <= n; i ++)
            p1[i] = rk[i], p2[i] = (i + bits[k] <= n) ? rk[i + bits[k]] : ;
        Rsort(p2, y, num); Rsort(p1, SA, y);
        for(int i = , p = ; i <= n; m = p, i ++)
            rk[SA[i]] = cmp(SA[i - ], SA[i]) ? p : ++ p;
        if(m >= n) break;
    }
}

void Get_Height()
{
    for(int i = , k = ; i <= n; i ++)
    {
        if(k) k --;
        int j = SA[rk[i] - ];
        while(max(i, j) + k -  <= n && a[j + k - ] == a[i + k - ]) k ++;
        ST[rk[i]][] = k;
    }
}

void Build()
{
    for(int j = ; j < CNST; j ++)
        for(int i = ; i + bits[j] -  <= n; i ++)
            ST[i][j] = min(ST[i][j - ], ST[i + bits[j - ]][j - ]);
}

int RMQ(int x, int y)
{
    if(y < x) swap(x, y);
    int k = Log[y - x + ];
    return min(ST[x][k], ST[y - bits[k] + ][k]);
}

int Query(int x, int len)
{
    int l = , r = rk[x], ans2 = rk[x], ans1 = rk[x] + ;
    while(l <= r)
    {
        int mid = (l + r) >> ;
        if(RMQ(rk[x], mid) >= len) ans1 = mid, r = mid - ;
        else l = mid + ;
    }

    ans1 --; l = rk[x] + , r = n;
    while(l <= r)
    {
        int mid = (l + r) >> ;
        if(RMQ(rk[x] + , mid) >= len) ans2 = mid, l = mid + ;
        else r = mid - ;
    }
    return ans2 - ans1 + ;
}

void init()
{
    Log[] = -; for(int i = ; i < maxn; i ++) Log[i] = Log[i >> ] + ;
    bits[] = ; for(int i = ; i < CNST; i ++) bits[i] = bits[i - ] << ;
    for(int i = ; i < maxn; i ++) num[i] = i;
}

int main()
{
    init();
    scanf("%s", a); n = strlen(a);
    pre(); manacher();
    for(int i = ; i <= n; i ++) rk[i] = a[i - ];
    Get_SA(); Get_Height(); Build();
    for(int i = ; i < N; i ++)
    {
        if(s[i] == '#') continue;
        int r = (i - ) >> ;
        int l = r - rec[i] + ; r ++, l ++;
        int t = Query(l, r - l + );
        ans = max(ans, 1ll * t * (r - l + ));
    }
    printf("%lld\n", ans);
    return ;
}