SPOJ694&&SPOJ705 DISUBSTR

来源：互联网发布：网络教育软件编辑：程序博客网时间：2024/06/03 19:26

泥萌都太强啦！！！

题意：
求一个串中本质不同的子串个数。

题解：
后缀数组论文题

# include <bits/stdc++.h># define N 50010class Suffix_Array  {private :    int n, a [N], b [N], c [N], sa [N], height [N], rank [N] ;public :    void build ( char* s, int n, int m )  {        int *x = a, *y = b, *t ;  this -> n = n ;        for ( register int i = 0 ; i < m ; c [i ++] = 0 ) ;        for ( register int i = 0 ; i < n ; ++ c [x [i] = s [i]], ++ i ) ;             for ( register int i = 1 ; i < m ; ++ i) c [i] += c [i - 1] ;        for ( register int i = n - 1 ; ~ i ; sa [-- c [x [i]]] = i, -- i );        for ( register int i, k = 1, p = 1 ; p < n && k <= n ; k <<= 1, m = p )  {            for ( p = 0, i = n - k ; i < n ; y [p ++] = i ++ ) ;            for ( i = 0 ; i < n ; sa [i] >= k ? y [p ++] = sa [i ++] - k : ++ i ) ;            for ( i = 0 ; i < m ; c [i ++] = 0 ) ;            for ( i = 0 ; i < n ; ++ c [x [y [i ++]]] ) ;             for ( i = 1 ; i < m ; ++ i )  c [i] += c [i - 1] ;            for ( i = n - 1 ; ~ i ; sa [-- c [x [y [i]]]] = y [i], -- i ) ;            for ( t = x, x = y, y = t, p = 1, x [sa [0]] = 0, i = 1 ; i < n ; ++ i )                x [sa [i]] = ( y [sa [i - 1]] == y [sa [i]] && y [sa [i - 1] + k] == y [sa [i] + k] ) ? p - 1 : p ++ ;        }        for ( register int i = 0 ; i < n ; rank [sa [i]] = i, ++ i ) ;        for ( register int i = 0, k = 1, j; i < n ; height [rank [i ++]] = k )            for ( k ? -- k : 0, j = sa [rank [i] - 1] ; s [i + k] == s [j + k] ; ++ k ) ;    }    inline long long QueryAns ( )  {        long long rt ( 0 ) ;        for ( int i = 1 ; i <= n - 1 ; ++ i )  rt += n - sa [i] - 1 - height  [i] ;        return rt ;    }} sa ;int main ( )  {    int T ;    scanf ( "%d", & T ) ;    while ( T -- )  {        static char s  [N] ;        scanf ( "%s", s ) ;        sa.build ( s, strlen ( s ) + 1, 500 ) ;        printf ( "%lld\n", sa.QueryAns ( ) ) ;    }}

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