POJ 3415 Common Substrings

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

2
aababaa
abaabaa
1
xx
xx
0

Sample Output

22

5

题目大意:
给定两串A，B，找出A，B的长度大于k的公共子串的个数

题解：
显然后缀数组，将两个串合并 然后显然中间加一个‘#’阻断height越界
然后会发现求出high之后两串若分属A，B两串 显然 对答案贡献为lcp-k+1
再根据老套路 利用 height 递增的性质去维护单调栈，记录栈中lcp-k+1的值的总和 tot
如果当前height[i]大于栈顶就统计 
如果height[i]<=栈顶 我们直接将栈顶和height[i]合并
调了很久最后发现把l开成了char = =|

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=400005;
int s[N];char S1[N],S2[N];int n,l,tmp[N],rk[N],sa[N],high[N],K,k;
bool comp(int i,int j){
    if(rk[i]!=rk[j])return rk[i]<rk[j];
    int ri=i+k<=n?rk[i+k]:-1;
    int rj=j+k<=n?rk[j+k]:-1;
    return ri<rj;
}
void Getsa(){
    for(int i=1;i<=n;i++){
        sa[i]=i;rk[i]=s[i];
    }
    for(k=1;k<=n;k<<=1){
        sort(sa+1,sa+n+1,comp);
        for(int i=1;i<=n;i++)tmp[sa[i]]=tmp[sa[i-1]]+comp(sa[i-1],sa[i]);
        for(int i=1;i<=n;i++)rk[i]=tmp[i];
    }
}
void Gethight(){
    int h=0,j;
    for(int i=1;i<=n;i++){
        j=sa[rk[i]-1];
        if(h)h--;
        for(;i+h<=n && j+h<=n;h++)if(s[i+h]!=s[j+h])break;
        high[rk[i]-1]=h;
    }
}
ll ans=0;
struct stacker{
    int lcp,cnt;
}st[N<<1];
void Getanswer()
{
    ll tot=0,cnt=0;int top=0;
    for(int i=1;i<n;i++){
       if(high[i]<K){
            top=tot=0;
            continue;
        }
        cnt=0;
        if(sa[i]<l)cnt++,tot+=high[i]-K+1;
        while(top && high[i]<=st[top].lcp){
            tot-=(ll)(st[top].lcp-high[i])*st[top].cnt;
            cnt+=st[top].cnt;
            top--;
        }
        st[++top].lcp=high[i];st[top].cnt=cnt;
        if(sa[i+1]>l)ans+=tot;
    }
    tot=0;top=0;cnt=0;
    for(int i=1;i<n;i++){
       if(high[i]<K){
            top=tot=0;
            continue;
        }
        cnt=0;
        if(sa[i]>l)cnt++,tot+=high[i]-K+1;
        while(top && high[i]<=st[top].lcp){
            tot-=(ll)(st[top].lcp-high[i])*st[top].cnt;
            cnt+=st[top].cnt;
            top--;
        }
        st[++top].lcp=high[i];st[top].cnt=cnt;
        if(sa[i+1]<l)ans+=tot;
    }
}
void work(){
    n=0;ans=0;
    scanf("%s%s",S1,S2);
    l=strlen(S1);
    for(int i=0;i<l;i++)s[++n]=S1[i];
    s[++n]='#';l++;
    for(int i=0,sz=strlen(S2);i<sz;i++)s[++n]=S2[i];
    s[n+1]=0;
    Getsa();
    Gethight();
    Getanswer();
    printf("%lld
",ans);
}
int main()
{
    freopen("pp.in","r",stdin);
    freopen("pp.out","w",stdout);
    while(scanf("%d",&K))
        {
            if(!K)break;
            work();
        }
    return 0;
}