传送门：【Tsinsen】A1393. Palisection
题目分析：首先串S倒着插入构造回文树，处理出以每个下标为结尾的回文串个数cnt2[now] = cnt2[fail[now]]+1，，然后处理出后缀和，suffix[i]表示开头下标大于等于下标i的回文串个数。
然后我们再正着插入字符构造回文树，每次插入结束后，ans+=cnt2[i]*suffix[i+1]。
当正着插入结束后，我们求得的ans恰是没有重叠的串的对数！所以我们只要再求得所有的对数，减去ans后就是答案。
代码如下：
#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;typedef long long LL ;#define rep( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; — i )#define clr( a , x ) memset ( a , x , sizeof a )const int MAXN = 2000005 ;const int mod = 51123987 ;const int N = 26 ;struct Palindromic_Tree {int next[MAXN][N] ;int fail[MAXN] ;int cnt2[MAXN] ;int cnt[MAXN] ;int len[MAXN] ;int S[MAXN] ;int last ;int n ;int p ;int newnode ( int l ) {rep ( i , 0 , N ) next[p][i] = 0 ;cnt[p] = 0 ;cnt2[p] = 0 ;len[p] = l ;return p ++ ;}void init () {p = 0 ;newnode ( 0 ) ;newnode ( -1 ) ;last = 0 ;n = 0 ;S[0] = -1 ;fail[0] = 1 ;}int get_fail ( int x ) {while ( S[n – len[x] – 1] != S[n] ) x = fail[x] ;return x ;}int add ( int c ) {c -= 'a' ;S[++ n] = c ;int cur = get_fail ( last ) ;if ( !next[cur][c] ) {int now = newnode ( len[cur] + 2 ) ;fail[now] = next[get_fail ( fail[cur] )][c] ;next[cur][c] = now ;cnt2[now] = cnt2[fail[now]] + 1 ;}last = next[cur][c] ;cnt[last] ++ ;return cnt2[last] ;}int count ( int ans = 0 ) {rev ( i , p – 1 , 1 ) {cnt[fail[i]] = ( cnt[fail[i]] + cnt[i] ) % mod ;ans = ( ans + cnt[i] ) % mod ;}return ans ;}} ;Palindromic_Tree T ;char s[MAXN] ;int suffix[MAXN] ;int n ;void solve () {int ans = 0 ;T.init () ;suffix[n] = 0 ;rev ( i , n – 1 , 0 ) suffix[i] = ( suffix[i + 1] + T.add ( s[i] ) ) % mod ;T.init () ;rep ( i , 0 , n ) ans = ( ans + ( LL ) T.add ( s[i] ) * suffix[i + 1] ) % mod ;int tot = T.count () ;tot = ( ( LL ) tot * ( tot – 1 ) / 2 ) % mod ;printf ( "%d\n" , ( ( tot – ans ) % mod + mod ) % mod ) ;}int main () {while ( ~scanf ( "%d%s" , &n , s ) ) solve () ;return 0 ;}
所有欺骗中，自欺是最为严重的
