hdu 3518 后缀数组

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后缀数组的题目主要是对三个数组的利用

sa[]：记录的是排名为i的后缀的首字母的下标（1 - n）

rank[]：记录的是首字母下标为i的后缀的排名（１－n )

height[] : 记录是j和j-1排序的最长公共前缀

这道题求取的是不重叠的最少出现两次的子串的个数

根据height[j]数组的性质，枚举长度然后按顺序遍历i，如果最长公共前缀大于枚举出的长度i，证明存在相同长度为i的不同子串，然后判断是否重叠即可，因为如果中间只要出现height[j]<i的情况，那么说明具有长度为i的相同子串的一组已经全部找到，判断当中首字母位置最靠后的和最靠前的差距是否大于相同串长度即可

#include <cstring>#include <algorithm>#include <cstdio>#include <iostream>#define MAX 1007using namespace std;char s[MAX];int r[MAX],sa[MAX],wa[MAX],wb[MAX],wv[MAX],wss[MAX],rank[MAX],height[MAX];int cmp ( int *r , int a , int b , int len ){    return r[a]==r[b]&&r[a+len]==r[b+len];}void da ( int* r , int* sa , int n , int m ){    int i,j,p, *x = wa , *y = wb , *t;    for ( i = 0 ; i < m ; i++ ) wss[i] = 0;    for ( i = 0 ; i < n ; i++ ) wss[x[i] = r[i]]++;    for ( i = 0 ; i < m ; i++ ) wss[i] += wss[i-1];    for ( i = n-1 ; i >= 0 ; i-- ) sa[--wss[x[i]]] = i;    for ( j = 1 , p = 1 ; p < n ; j *= 2 , m = p )    {        for ( p = 0 , i = n - j ; i < n ; i++ ) y[p++] = i;        for ( i = 0 ; i < n ; i++ ) if ( sa[i] >= j ) y[p++] = sa[i]-j;        for ( i = 0 ; i < n ; i++ ) wv[i] = x[y[i]];        for ( i = 0 ; i < m ; i++ ) wss[i] = 0;        for ( i = 0 ; i < n ; i++ ) wss[wv[i]]++;        for ( i = 0 ; i < m ; i++ ) wss[i] += wss[i-1];        for ( i = n-1 ; i >= 0 ; i-- ) sa[--wss[wv[i]]] = y[i];        for ( t = x , x = y , y = t , p = 1 , x[sa[0]] = 0 , i = 1 ; i < n ; i++ )            x[sa[i]] = cmp ( y , sa[i-1] , sa[i] , j ) ? p-1 : p++;    }}void calheight ( int *r , int *sa , int n ){    int i , j , k = 0;    for ( i = 1 ; i <= n ; i++ ) rank[sa[i]] = i;    for ( i = 0 ; i < n ; height[rank[i++]] = k )         for ( k?k--:0, j = sa[rank[i]-1] ; r[i+k] == r[j+k] ; k++ );}int main ( ){    while ( scanf ( "%s" , s ) )    {        if ( s[0] == '#' ) break;        int len = strlen ( s );        for ( int i = 0 ; i < len ; i++ ) r[i] = s[i] -'a' + 1;        r[len] = 0;        da ( r , sa , len+1 , 27 );        calheight ( r , sa , len );        int ans = 0;        for ( int i = 1 ; i <= (len+1)/2 ; i++ )        {            int maxn = -1 , minn = MAX;            for ( int j = 2 ; j <= len ; j++ )            {                if ( height[j] >= i )                {                    int temp = max ( sa[j] , sa[j-1] );                    maxn = max ( temp , maxn );                    temp = min ( sa[j] , sa[j-1] );                    minn = min ( temp , minn );                }                else                 {                    if ( maxn - minn >= i ) ans++;                    minn = MAX , maxn = -1;                }            }            if ( maxn - minn >= i ) ans++;        }        printf ( "%d\n" , ans );    }}

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