pku3415 Common Substrings

后缀数组+栈的线性扫描统计两个字符串的长度不少于K的公共子串个数

separate()构造原字符串str1和str2的高度数组，根据lcp(sa[i],sa[j])=rmq(height[i+1..j])，注意到各后缀在合并后得到的字符串中保持在原字符串中的字典序

 

 

#include  < iostream >
using   namespace  std;

#define  MAXN 200010
#define  clr(x) memset(x,0,sizeof(x))

int  b[MAXN],array[ 4 ][MAXN], * sa, * nsa, * rank, * nrank,
h1[MAXN],h2[MAXN],h3[MAXN],K,n1,n2,n3;
char  str1[MAXN / 2 ],str2[MAXN / 2 ],str3[MAXN];
int  st[MAXN];


void  make_sa(){
     int  i,j,k;

    sa = array[ 0 ];
    nsa = array[ 1 ];
    rank = array[ 2 ];
    nrank = array[ 3 ];

    memset(b, 0 , sizeof (b));
     for (i = 0 ;i < n3;i ++ )
        b[str3[i]] ++ ;
     for (i = 1 ;i <= 256 ;i ++ )
        b[i] += b[i - 1 ];
     for (i = n3 - 1 ;i >= 0 ;i -- )
        sa[ -- b[str3[i]]] = i;

     for (rank[sa[ 0 ]] = 0 ,i = 1 ;i < n3;i ++ ){
        rank[sa[i]] = rank[sa[i - 1 ]];
         if (str3[sa[i]] != str3[sa[i - 1 ]])
            rank[sa[i]] ++ ;
    }

     for (k = 1 ;k < n3  &&  rank[sa[n3 - 1 ]] < n3 - 1 ;k *= 2 ){
         for (i = 0 ;i < n3;i ++ )
            b[rank[sa[i]]] = i;
         for (i = n3 - 1 ;i >= 0 ;i -- )
             if (sa[i] - k >= 0 )
                nsa[b[rank[sa[i] - k]] -- ] = sa[i] - k;

         for (i = n3 - k;i < n3;i ++ )
            nsa[b[rank[i]] -- ] = i;

         for (nrank[nsa[ 0 ]] = 0 ,i = 1 ;i < n3;i ++ ){
            nrank[nsa[i]] = nrank[nsa[i - 1 ]];
             if (rank[nsa[i]] != rank[nsa[i - 1 ]]  ||  rank[nsa[i] + k] != rank[nsa[i - 1 ] + k])
                nrank[nsa[i]] ++ ;
        }

         int   * t = sa;sa = nsa;nsa = t;
        t = rank;rank = nrank;nrank = t;
    }

     for (i = 0 ,k = 0 ;i < n3;i ++ ){
         if (rank[i] == 0 )
            h3[rank[i]] = 0 ;
         else {
             for (j = sa[rank[i] - 1 ];str3[i + k] == str3[j + k];k ++ );
            h3[rank[i]] = k;
             if (k > 0 )
                k -- ;
        }
    }
}





__int64 solve( int  h[], int  n){
     int  i = 0 ,t,top = 0 ,m,num;
    __int64 cnt = 0 ;
    st[top] = 0 ;
    h[ 0 ] = h[n] = K - 1 ;
     while (i <= n){
        t = h[st[top]]; // 定义与height[st[top]]相关的后缀为栈顶后缀（乱编的）
         if (h[i] < K  &&  top == 0 )
            i ++ ;
         else
             if (h[i] > t)
                st[ ++ top] = i ++ ;
         else
             if (h[i] == t)
                i ++ ;
         else {
            m = i - st[top] + 1 ;
             if (h[i] >= K  &&  h[i] > h[st[top - 1 ]]){
                 // 统计与当前后缀和栈顶后缀的公共前缀不相关的
                 // 且属于栈顶后缀的公共前缀的部分（再读一遍？）
                num = t - h[i];
                h[st[top]] = h[i]; // 下次统计当前后缀和栈顶后缀的长度为h[i]的公共前缀部分
            }
             else // 当前后缀不能与栈顶后缀构成长度至少为K公共前缀，故暂不
                 // 纳入统计，且看它与下一个后缀能否构成公共K前缀
                num = t - h[st[ -- top]];
            cnt += (__int64)m * (m - 1 ) / 2 * num;
        }
    }
     return  cnt;
}

void  separate(){
    clr(h1);
    clr(h2);
     int  mn1 = INT_MAX,mn2 = INT_MAX,i,cnt1 = 0 ,cnt2 = 0 ;
     for (i = 0 ;i < n3;i ++ ){
         if (h3[i] < mn1)
            mn1 = h3[i];
         if (h3[i] < mn2)
            mn2 = h3[i];
         if (sa[i] < n1){
            h1[cnt1 ++ ] = mn1;
            mn1 = INT_MAX;
        }
         else {
            h2[cnt2 ++ ] = mn2;
            mn2 = INT_MAX;
        }
    }
}


int  main(){
    __int64 ans;
     while (scanf( " %d " , & K)  &&  K){
        scanf( " %s%s " ,str1,str2);
        n1 = strlen(str1);
        n2 = strlen(str2);
        str1[n1 ++ ] = ' # ' ;
        str1[n1] = ' \0 ' ;
        str2[n2 ++ ] = ' $ ' ;
        str2[n2] = ' \0 ' ;
        strcpy(str3,str1);
        strcat(str3,str2);
        n3 = n1 + n2;

        make_sa();
        separate();

        ans = solve(h3,n3) - solve(h2,n2) - solve(h1,n1);
        printf( " %I64d\n " ,ans);
    }
     return   0 ;
}