poj 3415 Common Substrings

Common Substrings

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3112		Accepted: 1002

Description

A substring of a string T is defined as:

 

T(i, k)=TiTi+1...Ti+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

 

/*<br />感謝qqz003大牛<br />經過一番折騰，總算是明白棧掃描這東西了<br />dc3還是比較犀利了，接著就刷進第一版了<br />*/<br />#include<cstdio><br />#include<cstring><br />const int maxn=202000;<br />int ws[maxn],wa[maxn],wb[maxn],wv[maxn];<br />int sa[maxn*5],height[maxn*5];<br />int rank[maxn],a[maxn];<br />char s[maxn];<br />int wstack[maxn][2];<br />#define F(x) ((x)/3+((x)%3==1?0:tb))<br />#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)<br />int c0(int *r,int a,int b)<br />{<br /> return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];<br />}<br />int c12(int k,int *r,int a,int b)<br />{<br /> if(k==2)<br /> return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);<br /> else<br /> return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];<br />}<br />void sort(int *r,int *a,int *b,int n,int m)<br />{<br /> int i;<br /> for(i=0; i<n; i++) wv[i]=r[a[i]];<br /> for(i=0; i<m; i++) ws[i]=0;<br /> for(i=0; i<n; i++) ws[wv[i]]++;<br /> for(i=1; i<m; i++) ws[i]+=ws[i-1];<br /> for(i=n-1; i>=0; i--) b[--ws[wv[i]]]=a[i];<br /> return;<br />}<br />void dc3(int *r,int *sa,int n,int m)<br />{<br /> int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;<br /> r[n]=r[n+1]=0;<br /> for(i=0; i<n; i++) if(i%3!=0) wa[tbc++]=i;<br /> sort(r+2,wa,wb,tbc,m);<br /> sort(r+1,wb,wa,tbc,m);<br /> sort(r,wa,wb,tbc,m);<br /> for(p=1,rn[F(wb[0])]=0,i=1; i<tbc; i++)<br /> rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;<br /> if(p<tbc) dc3(rn,san,tbc,p);<br /> else for(i=0; i<tbc; i++) san[rn[i]]=i;<br /> for(i=0; i<tbc; i++)<br /> if(san[i]<tb) wb[ta++]=san[i]*3;<br /> if(n%3==1) wb[ta++]=n-1;<br /> sort(r,wb,wa,ta,m);<br /> for(i=0; i<tbc; i++) wv[wb[i]=G(san[i])]=i;<br /> for(i=0,j=0,p=0; i<ta && j<tbc; p++)<br /> sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];<br /> for(; i<ta; p++) sa[p]=wa[i++];<br /> for(; j<tbc; p++) sa[p]=wb[j++];<br /> return;<br />}<br />void cal(int *r,int *sa,int n)<br />{<br /> int i,j,k=0;<br /> for (i=1; i<=n; i++) rank[sa[i]]=i;<br /> for (i=0; i<n; height[rank[i++]]=k)<br /> for (k?k--:0,j=sa[rank[i]-1]; r[i+k]==r[j+k]; k++);<br /> return;<br />}<br />int main()<br />{<br /> int k;<br /> int len1,len2,n;<br /> while(scanf("%d",&k)!=EOF)<br /> {<br /> if(k==0) break;<br /> scanf("%s",s);<br /> len1=strlen(s);<br /> for(int i=0;i<len1;++i) a[i]=(int)s[i];<br /> a[len1]=1;<br /> scanf("%s",s);<br /> len2=strlen(s);<br /> for(int i=0;i<len2;++i) a[len1+1+i]=(int)s[i];<br /> a[len1+len2+1]=0;<br /> n=len1+len2+1;<br /> dc3(a,sa,n+1,128);<br /> cal(a,sa,n);<br /> //棧掃描<br /> for(int i=1;i<=n;i++)//初始化，只需要滿足首碼大於k的即可<br /> {<br /> height[i]-=k-1;<br /> if(height[i]<0) height[i]=0;<br /> }<br /> int top=0;<br /> long long tsum=0,sum=0;<br /> int w;<br /> for(int i=1;i<=n;i++)<br /> {<br /> w=0;<br /> while(top>0&&wstack[top-1][0]>height[i])<br /> {<br /> top--;<br /> tsum+=(height[i]-wstack[top][0])*wstack[top][1];//更新<br /> w+=wstack[top][1];//壓縮<br /> }<br /> wstack[top][0]=height[i];//壓縮<br /> wstack[top++][1]=w;<br /> if(sa[i-1]<len1)//還在A串,就加入這個尾碼<br /> {<br /> tsum+=height[i];<br /> wstack[top-1][1]++;<br /> }<br /> if(sa[i]>len1)//現在討論B串，到了一個有B串的，把前面的加起來<br /> sum+=tsum;<br /> }</p><p> top=0;<br /> tsum=0;<br /> for(int i=1;i<=n;i++)<br /> {<br /> w=0;<br /> while(top>0&&wstack[top-1][0]>height[i])<br /> {<br /> top--;<br /> tsum+=(height[i]-wstack[top][0])*wstack[top][1];//更新<br /> w+=wstack[top][1];//壓縮<br /> }<br /> wstack[top][0]=height[i];//壓縮<br /> wstack[top++][1]=w;<br /> if(sa[i-1]>len1)//還在B串<br /> {<br /> tsum+=height[i];<br /> wstack[top-1][1]++;<br /> }<br /> if(sa[i]<len1)//現在討論A串，到了一個有A串的，把前面的加起來<br /> {<br /> sum+=tsum;<br /> }<br /> }<br /> printf("%lld/n",sum);<br /> }<br /> return 0;<br />}<br />