【题解】回文串 APIO 2014 BZOJ 3676 COGS 1985 Manacher+后缀数组+二分

这题可以用回文自动机来做，但是我并没有学，于是用Manacher+SA的做法O(nlogn)水过

首先，看到回文串就能想到用Manacher

同样还是要利用Manacher能不重复不遗漏地枚举每个回文子串的性质

只是不重复不遗漏还不够，我们还要统计出现次数

每个子串一定是一个后缀的前缀，于是可以用后缀数组

用后缀数组求出height数组之后，对于在Manacher过程中枚举到的每个长度为k的回文串，可以在height数组中二分，用O(logn)的时间求出这个子串的出现次数

BZOJ和COGS上有评论说Manacher + SA的方式被卡了，也有人说自己跑了19s，我这个实现是在BZOJ上跑了10s，COGS的76组数据总共跑了3.7s。

代码如下：

 #include <cstring>
 #include <algorithm>
 #include <cstdio>
 #include <cctype>

 using namespace std;
 typedef long long ll;
 const int MAXN = , LOGN = ;

 int n;
 char str[MAXN];
 int sas[MAXN], san;
 int mas[MAXN<<], man;

 namespace SA {
     int sa[MAXN], rk[MAXN], ht[MAXN];
     int tmp1[MAXN], tmp2[MAXN], cnt[MAXN];
     int minv[MAXN][LOGN], logn[MAXN];
     void solve( int m ) {
         int *x = tmp1, *y = tmp2;
         for( int i = ; i < m; ++i ) cnt[i] = ;
         for( int i = ; i < san; ++i ) ++cnt[ x[i] = sas[i] ];
         for( int i = ; i < m; ++i ) cnt[i] += cnt[i-];
         for( int i = san-; i >= ; --i ) sa[--cnt[x[i]]] = i;
         for( int k = ; k <= san; k <<=  ) {
             int p = ;
             for( int i = san-k; i < san; ++i ) y[p++] = i;
             for( int i = ; i < san; ++i ) if( sa[i] >= k ) y[p++] = sa[i]-k;
             for( int i = ; i < m; ++i ) cnt[i] = ;
             for( int i = ; i < san; ++i ) ++cnt[x[i]];
             for( int i = ; i < m; ++i ) cnt[i] += cnt[i-];
             for( int i = san-; i >= ; --i ) sa[--cnt[x[y[i]]]] = y[i];
             swap(x,y), x[sa[]] = , p = ;
             for( int i = ; i < san; ++i )
                 x[sa[i]] = y[sa[i]] == y[sa[i-]] && y[sa[i]+k] == y[sa[i-]+k] ? p- : p++;
             if( p == san ) break;
             m = p;
         }
         for( int i = ; i < san; ++i ) rk[i] = x[i];
         int k = ;
         for( int i = ; i < san; ++i ) {
             if( k ) --k;
             if( !rk[i] ) continue;
             int j = sa[rk[i]-];
             while( sas[i+k] == sas[j+k] ) ++k;
             ht[rk[i]] = minv[rk[i]][] = k;
         }
         for( int k = ; (<<k) <= san; ++k )
             for( int i = ; i+(<<k) <= san; ++i )
                 minv[i][k] = min( minv[i][k-], minv[i+(<<(k-))][k-] );
         k = ;
         for( int i = ; i <= san; ++i ) {
             if( (<<(k+)) <= i ) ++k;
             logn[i] = k;
         }
     }
     int qmin( int l, int r ) {
         int k = logn[r-l+];
         return min( minv[l][k], minv[r+-(<<k)][k] );
     }
 }

 void input() {
     scanf( "%s", str ), n = strlen(str);
     man = ;
     for( int i = ; i < n; ++i ) {
         sas[i] = str[i];
         mas[man++] = '#', mas[man++] = str[i];
     }
     sas[n] = , san = n+;
     mas[man++] = '#';
     SA::solve();
 }

 ll ans = ;
 int rd[MAXN<<];
 void update( int p, int k ) {
     using namespace SA;
     p = rk[p];
     int LL = , LR = p;
     while( LL < LR ) {
         int mid = (LL+LR)>>;
         if( qmin(mid+,p) >= k ) LR = mid;
         else LL = mid+;
     }
     int RL = p, RR = san-;
     while( RL < RR ) {
         int mid = (RL+RR+)>>;
         if( qmin(p+,mid) >= k ) RL = mid;
         else RR = mid-;
     }
     ans = max( ans, ll(k)*(RL-LL+) );
 }
 int cnt[] = {};
 void manacher() {
     int mx = , p = ;
     for( int i = ; i < man; ++i ) {
         if( i < mx ) rd[i] = min( rd[*p-i], mx-i );
         else rd[i] = ;
         while( i+rd[i] < man && i-rd[i] >=  && mas[i+rd[i]] == mas[i-rd[i]] ) {
             if( islower( mas[i-rd[i]] ) ) update( (i-rd[i])/, rd[i]+ );
             ++rd[i];
         }
         if( i+rd[i] > mx ) mx = i+rd[i], p = i;
     }
     for( int i = ; i < n; ++i )
         ans = max( ans, (ll)++cnt[str[i]-'a'] );
     printf( "%lld\n", ans );
 }

 int main() {
     // freopen( "apio2014_palindrome.in", "r", stdin );
     // freopen( "apio2014_palindrome.out", "w", stdout );
     input(), manacher();
     return ;
 }