SPOJ 705 Distinct Substrings（后缀数组）

SPOJ 705 Distinct Substrings（后缀数组）

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 <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Sample Input:

2
CCCCC
ABABA

Sample Output:

5
9

Hint

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

  给出一个字符串，输出其不同子串的个数。

解题思路

  如果直接枚举用set一个一个存下它的子串极有可能会超时，所以这里我们用后缀数组来求解。这题跟SPOJ694一样，就把那题的代码改个上限就可以了。在转long long时跪了一次，菜的安详orz。

代码

#include <cstdio>
#include <iostream>
#include <cstring>
#define  LL long long
#define  ULL unsigned long long
using namespace std;
const int maxn= 5e4+50;

int wa[maxn],wb[maxn],wv[maxn],Ws[maxn];

int cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(char r[],int sa[],int n,int m)
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++) Ws[i]=0;
    for(i=0; i<n; i++) Ws[x[i]=r[i]]++;
    for(i=1; i<m; i++) Ws[i]+=Ws[i-1];
    for(i=n-1; i>=0; i--) sa[--Ws[x[i]]]=i;
    for(j=1,p=1; 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++) Ws[i]=0;
        for(i=0; i<n; i++) Ws[wv[i]]++;
        for(i=1; i<m; i++) Ws[i]+=Ws[i-1];
        for(i=n-1; i>=0; i--) sa[--Ws[wv[i]]]=y[i];
        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
    return;
}
int sa[maxn],rankk[maxn],height[maxn],n;

void getheight(char *r,int *sa,int n)
{
    int i,j,k=0;
    for(i=1; i<=n; i++) rankk[sa[i]]=i;
    for(i=0; i<n; height[rankk[i++]]=k)
        for(k?k--:0,j=sa[rankk[i]-1]; r[i+k]==r[j+k]; k++);
    for(int i=n; i>=1; --i) ++sa[i],rankk[i]=rankk[i-1];
}

char arr[maxn];
int main()
{
#ifdef DEBUG
    freopen("in.txt","r",stdin);
#endif // DEBUG
    int t;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        scanf("%s",arr);
        int len=strlen(arr);
        da(arr,sa,len+1,130);
        getheight(arr,sa,len);
        LL ans=(LL)(len+1)*len/2;
        for(int i=1; i<=len-1; i++)
            ans-=height[i+1];
        printf("%lld\n",ans);
    }
    return 0;
}