spoj 1812 lcsII (后缀自动机)

spoj 1812 lcsII (后缀自动机)

题意：求多个串的lcs，最多10个串，每个串最长10w

解题思路：后缀自动机。先建好第一个串的sam，然后后面的串拿上去跑（这个过程同前一题）。sam上，节点要记录两个信息，先说mi[p]，其意思是p节点能匹配的所有串的最短的长度是多少，那我们如何更新这个mi呢？于是我们要记录另一个信息，mm[p]表示，跑当前串时，所跑到的节点能匹配的最大值，用这个最大值去更新mi。一个串跑完之后，要根据parent树，把mm往fa更新，然后用底下传上来的mm更新mi。怎么更新parent呢？很简单，我们知道，parent tree中父亲的代表串的长度必然比儿子短，所以我们只要根据代表串的长度排序，然后从后往前更新即可。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std ;const int maxn = 311111 ;int max ( int a , int b ) { return a > b ? a : b ; }int min ( int a , int b ) { return a < b ? a : b ; }int wv[maxn<<1] , ws[maxn<<1] ;int mm[maxn<<1] , mi[maxn<<1] ;struct sam {int fa[maxn<<1] , c[26][maxn<<1] , val[maxn<<1] , pos[maxn<<1] ;int tot , last ;inline int new_node ( int step ) {val[++tot] = step ;int i ;for ( i = 0 ; i < 26 ; i ++ ) c[i][tot] = 0 ;fa[tot] = 0 , mi[tot] = val[tot] ;return tot ;}void add ( int k ) {int p = last , i ;int np = new_node ( val[p] + 1 ) ;while ( p && !c[k][p] ) c[k][p] = np , p = fa[p] ;if ( !p ) fa[np] = 1 ;else {int q = c[k][p] ;if ( val[p] + 1 == val[q] ) fa[np] = q ;else {int nq = new_node ( val[p] + 1 ) ;for ( i = 0 ; i < 26 ; i ++ ) c[i][nq] = c[i][q] ;fa[nq] = fa[q] ;fa[q] = fa[np] = nq ;while ( p && c[k][p] == q ) c[k][p] = nq , p = fa[p] ;}}last = np ;}void build ( char *s ) {tot = 0 ;last = new_node ( 0 ) ;int len = strlen ( s ) , i ;for ( i = 0 ; i < len ; i ++ ) add ( s[i] - 'a' ) ;for ( i = 1 ; i <= tot ; i ++ ) wv[val[i]] ++ ;for ( i = 1 ; i <= tot ; i ++ ) wv[i] += wv[i-1] ;for ( i = 1 ; i <= tot ; i ++ ) ws[wv[val[i]]--] = i ;}void work ( char *s ) {int len = strlen ( s ) ;int p = 1 , i , temp = 0 ;for ( i = 0 ; i < len ; i ++ ) {int k = s[i] - 'a' ;if ( c[k][p] ) temp ++ , p = c[k][p] ;else {while ( p && !c[k][p] ) p = fa[p] ;if ( !p ) temp = 0 , p = 1 ;else temp = val[p] + 1 , p = c[k][p] ;}mm[p] = max ( mm[p] , temp ) ;}for ( i = tot ; i >= 1 ; i -- ) {p = ws[i] ;mi[p] = min ( mi[p] , mm[p] ) ;if ( fa[p] && mm[fa[p]] < mm[p] ) mm[fa[p]] = mm[p] ;mm[p] = 0 ;}}int solve () {int i , ans = 0 ;for ( i = 1 ; i <= tot ; i ++ )ans = max ( ans , mi[i]) ;return ans ;}} suf ;char s[maxn] ;int main () {//freopen ( "a.txt" , "r" , stdin ) ;scanf ( "%s" , s ) ;suf.build ( s ) ;int cnt = 0 ;while ( scanf ( "%s" , s ) != EOF ) {suf.work ( s ) ;//cnt ++ ;//if ( cnt == 2 ) break ;}printf ( "%d\n" , suf.solve () ) ;}/*skdsfkdsaajfaa*/