POJ_P3415 Common Substring(后缀数组+单调栈)

POJ传送门

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 9232 Accepted: 3066
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

Source

POJ Monthly–2007.10.06, wintokk

Sol
单调栈+后缀数组
论文题

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 200005
int t1[N],t2[N],c[N];
int sa[N],rk[N],ht[N];
int n,k,l1,l2;char s[N];
void DA(int m,int n){
    int *x=t1,*y=t2;
    for(int i=0;i<m;i++) c[i]=0;
    for(int i=0;i<n;i++) c[x[i]=s[i]]++;
    for(int i=1;i<m;i++) c[i]+=c[i-1];
    for(int i=n-1;~i;i--) sa[--c[x[i]]]=i;
    for(int k=1,p=0;k<=n;k<<=1,p=0){
        for(int i=n-k;i<n;i++) y[p++]=i;
        for(int i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(int i=0;i<m;i++) c[i]=0;
        for(int i=0;i<n;i++) c[x[y[i]]]++;
        for(int i=1;i<m;i++) c[i]+=c[i-1];
        for(int i=n-1;~i;--i) sa[--c[x[y[i]]]]=y[i];
        swap(x,y);m=1,x[sa[0]]=0;
        for(int i=1;i<n;i++) x[sa[i]]=y[sa[i]]==y[sa[i-1]]&&
            (y[sa[i]+k]==y[sa[i-1]+k])?m-1:m++;
        if(m>=n) break;
    }
}
void CalcHeight(int n){
    for(int i=0;i<=n;i++) rk[sa[i]]=i;
    for(int i=0,j,k=0;i<n;ht[rk[i++]]=k){
        if(!rk[i]) continue;
        j=sa[rk[i]-1];if(k) k--;
        while(s[i+k]==s[j+k]) k++;
    }
    ht[0]=0;
}
int stk1[N],stk2[N],t;long long ans;
void GetAns(){
    ans=0,t=0;long long tot=0,cnt;
    for(int i=1;i<=n;i++){
        if(ht[i]<k){tot=0,t=0;continue;}
        if(cnt=0,sa[i-1]<l1) tot+=ht[i]-k+1,++cnt;
        while(t&&ht[i]<=ht[stk1[t-1]])
            t--,tot-=stk2[t]*(ht[stk1[t]]-ht[i]),cnt+=stk2[t];
        stk1[t]=i,stk2[t++]=cnt;
        if(sa[i]>l1) ans+=tot;
    }
    t=0,tot=0;
    for(int i=1;i<=n;i++){
        if(ht[i]<k){tot=0,t=0;continue;}
        if(cnt=0,sa[i-1]>l1) tot+=ht[i]-k+1,++cnt;
        while(t&&ht[i]<=ht[stk1[t-1]])
            t--,tot-=stk2[t]*(ht[stk1[t]]-ht[i]),cnt+=stk2[t];
        stk1[t]=i,stk2[t++]=cnt;
        if(sa[i]<l1) ans+=tot;
    }
}
char str1[N],str2[N];
int main(){
    while(scanf("%d",&k),k){
        scanf("%s%s",str1,str2);n=0;
        for(int i=0;str1[i];i++) s[n++]=str1[i];
        l1=n,s[n++]=1;
        for(int i=0;str2[i];i++) s[n++]=str2[i];
        s[n]=0;DA(128,n+1),CalcHeight(n);GetAns();printf("%lld\n",ans);
    }
    return 0;
}