UVALive 6869 Repeated Substrings(后缀数组)

String analysis often arises in applications from biology and
chemistry, such as the study of DNA and protein molecules. One
interesting problem is to find how many substrings are repeated
(at least twice) in a long string.
In this problem, you will write a program to find the total
number of repeated substrings in a string of at most 100 000
alphabetic characters. Any unique substring that occurs more
than once is counted. As an example, if the string is “aabaab”,
there are 5 repeated substrings: “a”, “aa”, “aab”, “ab”, “b”.
If the string is “aaaaa”, the repeated substrings are “a”, “aa”, “aaa”, “aaaa”. Note that repeated
occurrences of a substring may overlap (e.g. “aaaa” in the second case).
Input
The first line contains a positive integer, specifying the number of cases to follow. Each of the following
line contains a nonempty string of up to 100 000 alphabetic characters.
Output
For each line of input, output one line containing the number of unique substrings that are repeated.
You may assume that the correct answer fits in a signed 32-bit integer.
Sample Input
3
aabaab
aaaaa
AaAaA
Sample Output
5
4
5

/*
后缀数组.
题意:求出现过两次以上的不同子串有多少种.
字串即字符串后缀的前缀.
ans+=ht[i]-ht[i-1].
用当前的贡献减去之前已经计算过的贡献.
然后把每个前缀的贡献累加.
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#define MAXN 100010
using namespace std;
int n,K,m=130,ans,sa[MAXN],rank1[MAXN],c[MAXN],ht[MAXN],t1[MAXN],t2[MAXN],s[MAXN];
char ch[MAXN];
bool cmp(int *y,int a,int b,int k)
{
    int a1=y[a],b1=y[b];
    int a2=a+k>=n?-1:y[a+k];
    int b2=b+k>=n?-1:y[b+k];
    return a1==b1&&a2==b2;
}
void slovesa()
{
    int *x=t1,*y=t2;
    for(int i=0;i<m;i++) c[i]=0;
    for(int i=0;i<n;i++) c[x[i]=s[i]]++;
    for(int i=1;i<m;i++) c[i]+=c[i-1];
    for(int i=n-1;i>=0;i--) sa[--c[x[i]]]=i;
    for(int k=1,p=0;k<=n;k<<=1,m=p,p=0)
    {
        for(int i=n-k;i<n;i++) y[p++]=i;
        for(int i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(int i=0;i<m;i++) c[i]=0;
        for(int i=0;i<n;i++) c[x[y[i]]]++;
        for(int i=1;i<m;i++) c[i]+=c[i-1];
        for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
        swap(x,y),p=1,x[sa[0]]=0;
        for(int i=1;i<n;i++)
        {
            if(cmp(y,sa[i-1],sa[i],k)) x[sa[i]]=p-1;
            else x[sa[i]]=p++;
        }
        if(p>=n) break;
    }
}
void sloveheight()
{
    int k=0;
    for(int i=0;i<n;i++) rank1[sa[i]]=i;
    for(int i=0;i<n;ht[rank1[i++]]=k)
    {
        int j=sa[rank1[i]-1];
        if(k) k--;
        while(j+k<n&&i+k<n&&s[i+k]==s[j+k]) k++;
    }
    ht[0]=0;
}
void slove()
{
    for(int i=1;i<n;i++)
    {
        int l=n-sa[i-1]-1;
        if(ht[i]>=ht[i-1]) ans+=ht[i]-ht[i-1];
    }
    printf("%d
",ans);
}
void Clear()
{
    ans=0,m=130;
    memset(sa,0,sizeof sa);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",ch);Clear();
        n=strlen(ch);
        for(int i=0;i<n;i++) s[i]=ch[i];
        s[n++]=0;
        slovesa(),sloveheight(),slove();
    }
    return 0;
}