SPOJ

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.

题意：

为字符串的子串个数

思路：

使用后缀数组解决。

按sa遍历后缀数组，每一个后缀的贡献即为n-sa[i]-lcp[i];

这里的lcp就是你们所说的height

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

#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a, x) cout<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define ls (t<<1)
#define rs ((t<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);

char s[maxn];
int len,Rank[maxn],sa[maxn],k,tmp[maxn];
bool compare_sa(int i, int j) {
    if (Rank[i] != Rank[j]) { return Rank[i] < Rank[j]; }
    //如果以i开始,长度为k的字符串的长度,已经超出了字符串尾,那么就赋值为-1
    //这是因为,在前面所有数据相同的情况下,字符串短的字典序小.
    int ri = i + k <= len ? Rank[i + k] : -inf;
    int rj = j + k <= len ? Rank[j + k] : -inf;
    return ri < rj;
}

void construct_sa() {
    //初始的RANK为字符的ASCII码
    for (int i = 0; i <= len; i++) {
        sa[i] = i;
        Rank[i] = i < len ? s[i] : -inf;
    }
    for (k = 1; k <= len; k *= 2) {
        sort(sa, sa + len + 1, compare_sa);
        tmp[sa[0]] = 0;
        //全新版本的RANK,tmp用来计算新的rank
        //将字典序最小的后缀rank计为0
        //sa之中表示的后缀都是有序的,所以将下一个后缀与前一个后缀比较,如果大于前一个后缀,rank就比前一个加一.
        //否则就和前一个相等.
        for (int i = 1; i <= len; i++) {
            tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
        }
        for (int i = 0; i <= len; i++) {
            Rank[i] = tmp[i];

        }
    }
}
int lcp[maxn];


void construct_lcp(){
//    for(int i=0;i<=n;i++){Rank[sa[i]]=i;}

    int h=0;
    lcp[0]=0;
    for(int i=0;i<len;i++){//i为后缀数组起始位置
        int j = sa[Rank[i]-1];//获取当前后缀的前一个后缀(排序后)
        if(h>0)h--;
        for(;j+h<len&&i+h<len;h++){
            if(s[j+h]!=s[i+h])break;
        }
        lcp[Rank[i]]=h;
    }
}


int main() {
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%s",s);
        len = strlen(s);
        construct_sa();
        construct_lcp();

        int ans=0;
        for(int i=1;i<=len;i++){
            ans+=(len-sa[i]-lcp[i]);
        }
        printf("%d
",ans);
    }

    return 0;
}