SPOJ 705（后缀数组求单个子串的不重复子串个数）

SPOJ Problem Set (classical) 705. New Distinct Substrings Problem code: SUBST1

 

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.

Example

Input:
2
CCCCC
ABABA

Output:
5
9

#include <stdio.h>
#include <string.h>
const int maxn = 500000*2;

#define F(x) ((x)/3+((x)%3==1?0:tb))  
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)   
int wa[maxn],wb[maxn],wv[maxn],ws[maxn],a[maxn],sa[maxn]; 
char str1[maxn];
int len1,sum; 
 
int c0(int *r,int a,int b)   
{return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];}   

int c12(int k,int *r,int a,int b)   
{if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);   
else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];}   

void sort(int *r,int *a,int *b,int n,int m)   
{   
    int i;   
    for(i=0;i<n;i++) wv[i]=r[a[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--) b[--ws[wv[i]]]=a[i];   
    return;   
}   
void dc3(int *r,int *sa,int n,int m)   
{   
    int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;   
    r[n]=r[n+1]=0;   
    for(i=0;i<n;i++) if(i%3!=0) wa[tbc++]=i;   
    sort(r+2,wa,wb,tbc,m);   
    sort(r+1,wb,wa,tbc,m);   
    sort(r,wa,wb,tbc,m);   
    for(p=1,rn[F(wb[0])]=0,i=1;i<tbc;i++)   
        rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;   
    if(p<tbc) dc3(rn,san,tbc,p);   
    else for(i=0;i<tbc;i++) san[rn[i]]=i;   
    for(i=0;i<tbc;i++) if(san[i]<tb) wb[ta++]=san[i]*3;   
    if(n%3==1) wb[ta++]=n-1;   
    sort(r,wb,wa,ta,m);   
    for(i=0;i<tbc;i++) wv[wb[i]=G(san[i])]=i;   
    for(i=0,j=0,p=0;i<ta && j<tbc;p++)   
        sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];   
    for(;i<ta;p++) sa[p]=wa[i++];   
    for(;j<tbc;p++) sa[p]=wb[j++];   
    return;   
}   

int rank[maxn],height[maxn];
void calheight(int *r,int *sa,int n)
{
	int i,j,k=0;
	for(i=1;i<=n;i++)rank[sa[i]]=i;
	for(i=0;i<n;height[rank[i++]]=k)
		for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
}

void solve()
{
	int i;
	sum = 0;
	for (i=1;i<=len1;i++)
		sum += len1 - sa[i] - height[i];
}


int main()
{
	int t,i;
	while (scanf("%d",&t)!=EOF)
	{
		while (t--)
		{
			scanf("%s",str1);
			len1 = strlen(str1);
			for (i=0;i<len1;i++)
			{
				a[i] = static_cast<int>(str1[i]);
			}
			a[len1] = 0;
			dc3(a,sa,len1+1,250);
			calheight(a,sa,len1);
			solve();
			printf("%d/n",sum);
		}
	}
	return 0;
}