題目鏈接:點擊打開鏈接
Boring counting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2617 Accepted Submission(s): 1062
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”).
AC代碼:
#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裏面,因此要加一個判斷。
}
printf("%d\n",ans);
}
}