CSU1632-Repeated Substrings

Repeated Substrings

Time Limit: 5 Sec   Memory Limit: 128 MB
Submit: 87   Solved: 35
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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：给出一个字符串，求出出现次数大于1的子串的有几种

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

const int N=200010;
const int INF=0x7FFFFFFF;

struct Sa
{
    char s[N];
    int rk[2][N],sa[N],h[N],w[N],now,n;
    int rmq[N][20],lg[N],bel[N];

    bool GetS()
    {
        scanf("%s",s+1);
        return true;
    }

    void getsa(int z,int &m)
    {
        int x=now,y=now^=1;
        for(int i=1; i<=z; i++) rk[y][i]=n-i+1;
        for(int i=1,j=z; i<=n; i++)
            if(sa[i]>z) rk[y][++j]=sa[i]-z;
        for(int i=1; i<=m; i++) w[i]=0;
        for(int i=1; i<=n; i++) w[rk[x][rk[y][i]]]++;
        for(int i=1; i<=m; i++) w[i]+=w[i-1];
        for(int i=n; i>=1; i--) sa[w[rk[x][rk[y][i]]]--]=rk[y][i];
        for(int i=m=1; i<=n; i++)
        {
            int *a=rk[x]+sa[i],*b=rk[x]+sa[i-1];
            rk[y][sa[i]]=*a==*b&&*(a+z)==*(b+z)?m-1:m++;
        }
    }

    void getsa(int m)
    {
        now=rk[1][0]=sa[0]=s[0]=0;
        n=strlen(s+1);
        for(int i=1; i<=m; i++) w[i]=0;
        for(int i=1; i<=n; i++) w[s[i]]++;
        for(int i=1; i<=m; i++) rk[1][i]=rk[1][i-1]+(bool)w[i];
        for(int i=1; i<=m; i++) w[i]+=w[i-1];
        for(int i=1; i<=n; i++) rk[0][i]=rk[1][s[i]];
        for(int i=1; i<=n; i++) sa[w[s[i]]--]=i;
        rk[1][n+1]=rk[0][n+1]=0;
        for(int x=1,y=rk[1][m]; x<=n&&y<=n; x<<=1) getsa(x,y);
        for(int i=1,j=0; i<=n; h[rk[now][i++]]=j?j--:j)
        {
            if(rk[now][i]==1) continue;
            int k=n-max(sa[rk[now][i]-1],i);
            while(j<=k&&s[sa[rk[now][i]-1]+j]==s[i+j]) ++j;
        }
    }

    void getrmq()
    {
        h[n+1]=h[1]=lg[1]=0;
        for(int i=2; i<=n; i++)
            rmq[i][0]=h[i],lg[i]=lg[i>>1]+1;
        for(int i=1; (1<<i)<=n; i++)
        {
            for(int j=2; j<=n; j++)
            {
                if(j+(1<<i)>n+1) break;
                rmq[j][i]=min(rmq[j][i-1],rmq[j+(1<<i-1)][i-1]);
            }
        }
    }

    int lcp(int x,int y)
    {
        int l=min(rk[now][x],rk[now][y])+1,r=max(rk[now][x],rk[now][y]);
        return min(rmq[l][lg[r-l+1]],rmq[r-(1<<lg[r-l+1])+1][lg[r-l+1]]);
    }

    void work()
    {
        getsa(300);
        int ans=h[2];
        for(int i=3;i<=n;i++)
            if(h[i]>h[i-1])
                ans+=h[i]-h[i-1];
        printf("%d\n",ans);
    }

} sa;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        sa.GetS();
        sa.work();
    }
    return 0;
}