poj3415之长度不小于k的公共子串个数
Common Substrings
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 6254 | Accepted: 2069 |
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
2 aababaa abaabaa 1 xx xx 0
Sample Output
22 5
#include<iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef 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;
}