POJ3415:Common Substrings(后缀数组+单调栈)

Description

A substring of a string T is defined as:

T(i, k)=T_iT_i+1...T_i+k-1, 1≤i≤i+k-1≤|T|.

Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):

S = {(i, j, k) | k≥K, A(i, k)=B(j, k)}.

You are to give the value of |S| for specific A, B and K.

Input

The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.

1 ≤ |A|, |B| ≤ 10⁵
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.

Output

For each case, output an integer |S|.

Sample Input

2aababaaabaabaa1xxxx0

Sample Output

225

Source

POJ Monthly--2007.10.06, wintokk

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define LS 2*i#define RS 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N (2*100000+10)#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8int wa[N],wb[N],wm[N],wv[N],sa[N];int *rank,height[N],s[N],a[N];//sa:字典序中排第i位的起始位置在str中第sa[i]//rank:就是str第i个位置的后缀是在字典序排第几//height：字典序排i和i-1的后缀的最长公共前缀bool cmp(int *r,int a,int b,int l){    return r[a] == r[b] && r[a+l] == r[b+l];}void getsa(int *r,int *sa,int n,int m){    int *x=wa,*y=wb,*t;    for(int i=0; i<m; ++i)wm[i]=0;    for(int i=0; i<n; ++i)wm[x[i]=r[i]]++;    for(int i=1; i<m; ++i)wm[i]+=wm[i-1];    for(int i=n-1; i>=0; --i)sa[--wm[x[i]]]=i;    for(int i=0,j=1,p=0; p<n; j=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<m; ++i)wm[i]=0;        for(i=0; i<n; ++i)wm[x[y[i]]]++;        for(i=1; i<m; ++i)wm[i]+=wm[i-1];        for(i=n-1; i>=0; --i)sa[--wm[x[y[i]]]]=y[i];        for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0; i<n; ++i)        {            x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;        }    }    rank=x;}void getheight(int *r,int *sa,int n){    for(int i=0,j=0,k=0; i<n; height[rank[i++]]=k)    {        for(k?--k:0,j=sa[rank[i]-1]; r[i+k] == r[j+k]; ++k);    }}int k;char s1[N];int len1;LL solve(int n,int len,int k){    int *mark=wa,*sta=wb,top=0,i;    LL sum=0,num[3]= {0};    for(i = 1;i<=n;i++)    {        if(height[i]<k)        {            top = num[1] = num[2] =0;        }        else        {            for(int size = top; size&&sta[size]>height[i]-k+1; size--)            {                num[mark[size]] += height[i]-k+1-sta[size];                sta[size] = height[i]-k+1;            }            sta[++top] = height[i]-k+1;            if(sa[i-1]<len) mark[top] = 1;            if(sa[i-1]>len) mark[top] = 2;            num[mark[top]]+=height[i]-k+1;            if(sa[i]<len) sum+=num[2];            if(sa[i]>len) sum+=num[1];        }    }    return sum;}int main(){    int i,j;    while(~scanf("%d",&k),k)    {        scanf("%s",s1);        int n = 0;        for(n = 0;s1[n]!='\0';n++)            s[n] = s1[n];        s[len1=n] = '#';        scanf("%s",s1+n+1);        n++;        for(;s1[n]!='\0';n++)            s[n] = s1[n];        s[n] = 0;        getsa(s,sa,n+1,201);        getheight(s,sa,n);        printf("%lld\n",solve(n,len1,k));    }    return 0;}