【codeforces】2014 Asia Xian Regional Contest G The Problem to Slow Down You 【Palindromic Tree】

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传送门：【codeforces】2014 Asia Xian Regional Contest G The Problem to Slow Down You 【Palindromic Tree】

题目分析：我们对两个字符串分别建立回文树，然后分别从节点0和1开始dfs，如果两个串都可以往某一个走，则ans+=两个串该节点下cnt的乘积，然后沿着这个方向继续dfs。

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 = 200005 ;const int N = 26 ;struct Palindromic_Tree {int next[MAXN][N] ;int fail[MAXN] ;int cnt[MAXN] ;int len[MAXN] ;int S[MAXN] , n ;int last ;int p ;int newnode ( int l ) {rep ( i , 0 , N ) next[p][i] = 0 ;cnt[p] = 0 ;len[p] = l ;return p ++ ;}void init () {p = 0 ;newnode (  0 ) ;newnode ( -1 ) ;last = 0 ;n = 0 ;S[n] = -1 ;fail[0] = 1 ;}int get_fail ( int x ) {while ( S[n - len[x] - 1] != S[n] ) x = fail[x] ;return x ;}void 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 ;}last = next[cur][c] ;cnt[last] ++ ;}void count () {rev ( i , p - 1 , 0 ) cnt[fail[i]] += cnt[i] ;}} ;Palindromic_Tree T1 , T2 ;char s1[MAXN] , s2[MAXN] ;int n1 , n2 ;LL ans ;void dfs ( int u , int v ) {rep ( i , 0 , 26 ) {int x = T1.next[u][i] , y = T2.next[v][i] ;if ( x && y ) {ans += ( LL ) T1.cnt[x] * T2.cnt[y] ;dfs ( x , y ) ;}}}void solve () {ans = 0 ;T1.init () ;T2.init () ;scanf ( "%s%s" , s1 , s2 ) ;n1 = strlen ( s1 ) ;n2 = strlen ( s2 ) ;rep ( i , 0 , n1 ) T1.add ( s1[i] ) ;rep ( i , 0 , n2 ) T2.add ( s2[i] ) ;T1.count () ;T2.count () ;dfs ( 0 , 0 ) ;dfs ( 1 , 1 ) ;printf ( "%I64d\n" , ans ) ;}int main () {int T ;scanf ( "%d" , &T ) ;For ( cas , 1 , T ) {printf ( "Case #%d: " , cas ) ;solve () ;}return 0 ;}

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