CSU-1632 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

题目大意：统计字符串中重复出现的子串数目。
题目分析：sum（max(height(i)-height(i-1),0)）即为答案。

代码如下：

//# define AC

# ifndef AC

# include<iostream>
# include<cstdio>
# include<cstring>
# include<vector>
# include<queue>
# include<list>
# include<cmath>
# include<set>
# include<map>
# include<string>
# include<cstdlib>
# include<algorithm>
using namespace std;
# define mid (l+(r-l)/2)

typedef long long LL;
typedef unsigned long long ULL;

const int N=100000;
const int mod=1e9+7;
const int INF=0x7fffffff;
const LL oo=0x7fffffffffffffff;

int SA[N+5];
int tSA[N+5];
int cnt[N+5];
int rk[N+5];
int *x,*y;
int height[N+5];

int idx(char c)
{
    if('a'<=c&&c<='z') return c-'a';
    return c-'A'+26;
}

bool same(int i,int j,int k,int n)
{
    if(y[i]-y[j]) return false;
    if(i+k<n&&j+k>=n) return false;
    if(i+k>=n&&j+k<n) return false;
    return y[i+k]==y[j+k];
}

void buildSA(char *s)
{
    int n=strlen(s);
    int m=52;
    x=rk,y=tSA;
    for(int i=0;i<m;++i) cnt[i]=0;
    for(int i=0;i<n;++i) ++cnt[x[i]=idx(s[i])];
    for(int i=1;i<m;++i) cnt[i]+=cnt[i-1];
    for(int i=n-1;i>=0;--i) SA[--cnt[x[i]]]=i;

    for(int k=1;k<=n;k<<=1){
        int 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) cnt[i]=0;
        for(int i=0;i<n;++i) ++cnt[x[y[i]]];
        for(int i=1;i<m;++i) cnt[i]+=cnt[i-1];
        for(int i=n-1;i>=0;--i) SA[--cnt[x[y[i]]]]=y[i];

        p=1;
        swap(x,y);
        x[SA[0]]=0;
        for(int i=1;i<n;++i)
            x[SA[i]]=same(SA[i],SA[i-1],k,n)?p-1:p++;
        if(p>=n) break;
        m=p;
    }
}

void getHeight(char *s)
{
    int n=strlen(s);
    for(int i=0;i<n;++i) rk[SA[i]]=i;
    int k=0;
    for(int i=0;i<n;++i){
        if(rk[i]==0){
            height[rk[i]]=k=0;
        }else{
            if(k) --k;
            int j=SA[rk[i]-1];
            while(i+k<n&&j+k<n&&s[i+k]==s[j+k])
                ++k;
            height[rk[i]]=k;
        }
    }
}

char str[N+5];

void solve()
{
    int n=strlen(str);
    int ans=0;
    for(int i=0;i<n;++i){
        if(height[i]>height[i-1])
            ans+=height[i]-height[i-1];
    }
    printf("%d
",ans);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str);
        buildSA(str);
        getHeight(str);
        solve();
    }
    return 0;
}

# endif