HDU 3518 Boring counting（后缀数组入门题）

题目链接：点击打开链接

Boring counting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2617 Accepted Submission(s): 1062



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

题意：求至少出现两次的不重叠子串的数目总和。后缀数组入门，但是我不会告诉你我wa了多少次。注意的地方写在代码里。

AC代码：

#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++;//可能从来没有进入过else里面，因此要加一个判断。

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