poj3415之长度不小于k的公共子串个数

Common Substrings

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 6254 Accepted: 2069

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

#include<iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <map>#include <iomanip>#define INF 99999999typedef long long LL;using namespace std;const int MAX=2*100000+10;int *rank,r[MAX],sa[MAX],height[MAX];int wa[MAX],wb[MAX],wm[MAX];char s[MAX];bool cmp(int *r,int a,int b,int l){return r[a] == r[b] && r[a+l] == r[b+l];}void makesa(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 calheight(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);}}LL calculate(int n,int len,int k){int *mark=wb,*ans=wm,Top=0;//num[1],num[2]分别表示字符串A,B,suffix(0~i-1)和suffix(i)的最长公共子串>=k的总个数LL sum=0,num[3]={0};for(int i=1;i<=n;++i){if(height[i]<k){Top=num[1]=num[2]=0;}else{for(int size=Top;size && ans[size]>height[i]-k+1;--size){//维护单调栈,ans记录的是suffix(j)和suffix(i-1)>=k的最长公共子串的个数,个数越多表示height[j]越大num[mark[size]]+=height[i]-k+1-ans[size];//suffix(j)和suffix(i)>=k的最长公共子串只能是长度为k~height[i],所以需要减去(ans[size]-(height[i]-k+1))ans[size]=height[i]-k+1;//更新个数(更新单调栈,使栈里面元素非递减)}ans[++Top]=height[i]-k+1;if(sa[i-1]<len)mark[Top]=1;//由于num新增加的结果是suffix(i-1)和suffix(i)的结果,所以是判断sa[i-1] if(sa[i-1]>len)mark[Top]=2;num[mark[Top]]+=height[i]-k+1;//增加由suffix(i-1)和suffix(i)产生的结果 if(sa[i]<len)sum+=num[2];//表示和suffix(i)产生的结果新增加B串的suffix(0~i-1)和suffix(i)>=k的个数if(sa[i]>len)sum+=num[1];//表示和suffix(i)产生的结果新增加A串的suffix(0~i-1)和suffix(i)>=k的个数}}return sum;}int main(){int k,n,len;while(~scanf("%d",&k),k){scanf("%s",s);for(n=0;s[n] != '\0';++n)r[n]=s[n];r[len=n]='#';scanf("%s",s+n+1);for(++n;s[n] != '\0';++n)r[n]=s[n];r[n]=0;makesa(r,sa,n+1,256);calheight(r,sa,n);cout<<calculate(n,len,k)<<endl;}return 0;}