spojDistinct Substrings【后缀数组 不重复子串】

Description
Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20;

Each test case consists of one string, whose length is <= 1000

Output

For each test case output one number saying the number of distinct substrings.

Example

Sample Input:

2

CCCCC

ABABA

Sample Output:

5

9

Explanation for the testcase with string ABABA:

len=1 : A,B

len=2 : AB,BA

len=3 : ABA,BAB

len=4 : ABAB,BABA

len=5 : ABABA

Thus, total number of distinct substrings is 9.
这个也是为数不多看懂做法并且明白的题之一==

如果不算重复的个数，长度为n的字符串可以组成的子串个数是n+(n-1)……+1个 ，（从一开始枚举），联想length[]的定义：排名相近的两个子串的最长公共前缀长度，那既然都是公共了，就多算了这些子串==

/*******************
spoj694. Distinct Substrings
2016.2.22
3789	0
C++ (g++ 4.9.2)
1678
*******************/
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 10005
int t1[MAXN],t2[MAXN],c[MAXN];
bool cmp(int *r,int a,int b,int l)
{
return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(int str[],int sa[],int rank[],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++) rank[sa[i]]=i;
for(i=0;i<n;i++)
{
if(k) k--;
j=sa[rank[i]-1];
while(str[i+k]==str[j+k]) k++;
height[rank[i]]=k;
}
}
char str[MAXN];
int r[MAXN],sa[MAXN],rank[MAXN],height[MAXN];
int main()
{
//  freopen("cin.txt","r",stdin);
int t;
scanf("%d",&t);
while(~scanf("%s",str))
{
int n=strlen(str);
for(int i=0;i<n;i++) r[i]=str[i];
r
=0;
da(r,sa,rank,height,n,256);
int ans=0;
for(int i=1;i<=n;i++)
{
ans+=(n-i+1-height[i]);
// printf("h=%d   ",height[i]);
}
printf("%d\n",ans);
}
return 0;
}