SPOJ 题目705 New Distinct Substrings（后缀数组，求不同的子串个数）

SUBST1 - New 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 <= 50000

Output

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

Example

Input:
2
CCCCC
ABABA

Output:
5
9

ac代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
using namespace std;
char str[50010];
int sa[50100],Rank[50100],rank2[50100],height[50010],c[50100],*x,*y,s[50100];
int n;
void cmp(int n,int sz)
{
	int i;
	memset(c,0,sizeof(c));
	for(i=0;i<n;i++)
		c[x[y[i]]]++;
	for(i=1;i<sz;i++)
		c[i]+=c[i-1];
	for(i=n-1;i>=0;i--)
		sa[--c[x[y[i]]]]=y[i];
}
void build_sa(int *s,int n,int sz)
{
	x=Rank,y=rank2;
	int i,j;
	for(i=0;i<n;i++)
		x[i]=s[i],y[i]=i;
	cmp(n,sz);
	int len;
	for(len=1;len<n;len<<=1)
	{
		int yid=0;
		for(i=n-len;i<n;i++)
		{
			y[yid++]=i;
		}
		for(i=0;i<n;i++)
			if(sa[i]>=len)
				y[yid++]=sa[i]-len;
			cmp(n,sz);
		swap(x,y);
		x[sa[0]]=yid=0;
		for(i=1;i<n;i++)
		{
			if(y[sa[i-1]]==y[sa[i]]&&sa[i-1]+len<n&&sa[i]+len<n&&y[sa[i-1]+len]==y[sa[i]+len])
				x[sa[i]]=yid;
			else
				x[sa[i]]=++yid;
		}
		sz=yid+1;
		if(sz>=n)
			break;
	}
	for(i=0;i<n;i++)
		Rank[i]=x[i];
}
void getHeight(int *s,int n)
{
	int k=0;
	for(int i=0;i<n;i++)
	{
		if(Rank[i]==0)
			continue;
		k=max(0,k-1);
		int j=sa[Rank[i]-1];
		while(s[i+k]==s[j+k])
			k++;
		height[Rank[i]]=k;
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i;
		scanf("%s",str);
		int len=strlen(str);
		for(i=0;i<len;i++)
			s[i]=str[i];
		s[len]=0;
		build_sa(s,len+1,260);
		getHeight(s,len);
		int sum=0;
		for(i=1;i<=len;i++)
		{
			sum+=len-sa[i]-height[i];
		}
		printf("%d\n",sum);
	}
}

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