Boring counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2617    Accepted Submission(s): 1062


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 題意：求至少出現兩次的不重疊子串的數目總和。字尾陣列入門，但是我不會告訴你我wa了多少次。注意的地方寫在程式碼裡。

AC程式碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;
int wa[200005],wb[200005],wv[200005],ws[200005];
int r[200005],height[200005],rank1[200005],sa[200005];
int cmp(int *r,int a,int b,int l){
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(int *r,int *sa,int n,int m){
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0;i<m;i++) ws[i]=0;
    for(i=0;i<n;i++) ws[x[i]=r[i]]++;
    for(i=1;i<m;i++) ws[i]+=ws[i-1];
    for(i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;
    for(j=1,p=1;p<n;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<n;i++) wv[i]=x[y[i]];
        for(i=0;i<m;i++) ws[i]=0;
        for(i=0;i<n;i++) ws[wv[i]]++;
        for(i=1;i<m;i++) ws[i]+=ws[i-1];
        for(i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];
        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
    return;
}
void calheight(int *r,int *sa,int n){
    int i,j,k=0;
    for(i=1;i<=n;i++) rank1[sa[i]]=i;
    for(i=0;i<n;height[rank1[i++]]=k)
        for(k?k--:0,j=sa[rank1[i]-1];r[i+k]==r[j+k];k++);
    return;
}
int main()
{
    int i,j,len;char s[1100];
    while(scanf("%s",s)!=EOF&&s[0]!='#')
    {
       len=strlen(s);
       for(i=0;i<len;i++)
       r[i]=s[i];
       r[len]=0;
       da(r,sa,len+1,300);
       calheight(r,sa,len);
       int Max,Min,ans=0;
       for(i=1;i<=len;i++)
       {
         Max=sa[1],Min=sa[1];
         for(j=2;j<=len;j++)
         {
            if(height[j]>=i)
            {
              Max=max(Max,max(sa[j-1],sa[j]));
              Min=min(Min,min(sa[j-1],sa[j]));
            }
            else
            {
              if(Max-Min>=i) ans++;
              Max=sa[j],Min=sa[j];
            }
         }
        if(Max-Min>=i) ans++;//可能從來沒有進入過else裡面，因此要加一個判斷。

       }
       printf("%d\n",ans);
    }
}