hdu 3518 Boring counting 后缀数组

Problem Description

035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.
Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).

Input

The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).

Output

For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.

Sample Input

aaaa
ababcabb
aaaaaa
#

Sample Output

2
3
3

字符串长度只有1000 考虑n^2，根据最长不可重叠子串，可以知道 每两个串之间的lcp是递减的，所以暴力跑递减，直到<=height[i-1] 退出，因为height[i-1]都是相当于判断过的，我们只需要判断的是 k-height[i-1]的就可以。

#include <bits/stdc++.h>using namespace std;const int MAXN = 1100;int t1[MAXN],t2[MAXN],c[MAXN];int n;bool cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(char str[],int sa[],int ra[],int height[],int n,int m){    n++;    int i,j,p,*x=t1,*y=t2;    for(i=0;i<m;i++) c[i]=0;    for(i=0;i<n;i++) c[x[i]=str[i]]++;    for(i=1;i<m;i++) c[i]+=c[i-1];    for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;    for(j=1;j<=n;j<<=1)    {        p=0;        for(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++) c[i]=0;        for(i=0;i<n;i++) c[x[y[i]]]++;        for(i=1;i<m;i++) c[i]+=c[i-1];        for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];        swap(x,y);        p=1;x[sa[0]]=0;        for(i=1;i<n;i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;        if(p>=n) break;        m=p;    }    int k=0;    n--;    for(i=0;i<=n;i++) ra[sa[i]]=i;    for(i=0;i<n;i++)     {        if(k) k--;        j=sa[ra[i]-1];        while(str[i+k]==str[j+k])k++;        height[ra[i]]=k;    }}int ra[MAXN],height[MAXN];int sa[MAXN],num[MAXN];typedef long long ll;char str[MAXN];int main(){    while(~scanf("%s",str))    {        if(str[0]=='#') break;        int n=strlen(str);        da(str,sa,ra,height,n,130);         ll ans=0;        for(int i=1;i<=n;i++)        {            int k=height[i];            int minn=min(sa[i],sa[i-1]),maxx=max(sa[i],sa[i-1]);            int j=i,an=0;            while(k&&j<=n)            {                       if(maxx-minn>=k)                 {                    ans+=max(k-height[i-1],0);                    break;                }                k=min(k,height[j+1]);                j++,maxx=max(maxx,sa[j]),minn=min(minn,sa[j]);            }        }        printf("%lld\n",ans );    }}