CSU1632: Repeated Substrings(后缀数组)

Description

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 input consists of at most 10 cases. 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

HINT

Source

题意：求字符串中所有出现至少2次的子串个数

思路：就拿第一组来说

后缀按字典序来排就是

aab

aabaab

ab

abaab

b

baab

那么height数组就是0，3，1，2，0，1

我们可以知道，对于一段升序中，这里没有出现过的子串就是maxn-minn，那么加起来即可

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int rank1[N],height[N],s[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
int cmp(int *r,int a,int b,int k)
{
    return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)//n要包含末尾添加的0
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++)  wsf[i]=0;
    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;
    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;
    p=1;
    j=1;
    for(; 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++)  wsf[i]=0;
        for(i=0; i<n; i++)  wsf[wv[i]]++;
        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];
        t=x;
        x=y;
        y=t;
        x[sa[0]]=0;
        for(p=1,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
    }
}
void getheight(int *r,int n)//n不保存最后的0
{
    int i,j,k=0;
    for(i=1; i<=n; i++)  rank1[sa[i]]=i;
    for(i=0; i<n; i++)
    {
        if(k)
            k--;
        else
            k=0;
        j=sa[rank1[i]-1];
        while(r[i+k]==r[j+k])
            k++;
        height[rank1[i]]=k;
    }
}

char str[N];

int main()
{
    int t,i,j,n,len;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",str);
        len = strlen(str);
        for(i = 0; i<len; i++)
            s[i] = str[i];
        s[len] = 0;
        getsa(s,sa,len+1,200);
        getheight(s,len);
        int cnt = 0,minn = 0,maxn = 0;
        for(i = 1; i<=len; i++)
        {
            if(height[i]>maxn)
            {
                maxn = height[i];
            }
            else
            {
                cnt+=maxn-minn;
                maxn = height[i];
                minn = height[i];
            }
        }
        cnt+=maxn-minn;
        printf("%d\n",cnt);
    }

    return 0;
}