SPOJ

DISUBSTR - Distinct Substrings

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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.

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题意：给一个字符串，求其不同的子串的个数。

思路：每一个子串都可以看作是后缀的前缀，对于一个后缀，首先会贡献后缀长度个子串，减去其高度数组的值即重复计算的部分。

/*SPOJ - DISUBSTR 多少个不同的子串求一个串中有多少个不同的子串每个子串一定是某个后缀的前缀，那么原问题等价于求所有后缀之间的不相同的前缀的个数。height表示的公共前缀的长度 = 相同串的个数所以  总数-sum(height[])即可*/#include <algorithm>#include <cmath>#include <queue>#include <iostream>#include <cstring>#include <map>#include <cstdio>#include <vector>#include <functional>#define lson (i<<1)#define rson ((i<<1)|1)using namespace std;typedef long long ll;const int maxn = 50005;int t1[maxn],t2[maxn],c[maxn];bool cmp(int *r,int a,int b,int l)  {    return r[a]==r[b] &&r[l+a] == r[l+b];}void get_sa(int str[],int sa[],int Rank[],int height[],int n,int m){    n++;    int p,*x=t1,*y=t2;    for(int i = 0; i < m; i++) c[i] = 0;    for(int i = 0; i < n; i++) c[x[i] = str[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 j = 1; j <= n; j <<= 1)  {        p = 0;        for(int i = n-j; i < n; i++) y[p++] = i;        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;        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++)            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)? p-1:p++;        if(p >= n) break;        m = p;    }    int k = 0;    n--;    for(int i = 0; i <= n; i++)        Rank[sa[i]] = i;    for(int i = 0; i < n; i++)  {        if(k) k--;        int j = sa[Rank[i]-1];        while(str[i+k] == str[j+k]) k++;        height[Rank[i]] = k;    }}int Rank[maxn],height[maxn];int sa[maxn];char str[maxn];int a[maxn];int n;int main(){    int T;    scanf("%d",&T);    while(T--)    {        int tot = 0;        scanf("%s",str);        int len = strlen(str);        for(int i = 0;i < len; i++)            a[i] = str[i];        a[len] = 0;        get_sa(a,sa,Rank,height,len,200);        long long res=0;        for(int i=1;i<=len;i++){            res+=len-sa[i]-height[i];        }        printf("%lld\n",res);    }}