spoj694/705 Distinct Substrings

题目链接：http://acm.hust.edu.cn/vjudge/problem/19282

题目大意：求不同子串的个数

解题思路：后缀数组..
suffix(i)对子串个数所做的贡献为len-sa[i]+1，因为要求要不同的，所以减去与它串重复的子串个数h[i]。即每个后缀对答案的贡献为len-sa[i]+1-h[i];(***)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
#define maxn 1100

int sa[maxn],rk[maxn],y[maxn],len;
int Rsort[maxn],wr[maxn],h[maxn];
char s[maxn];
bool cmp(int x,int y,int ln){return (wr[x]==wr[y])&&(wr[x+ln]==wr[y+ln]);}
void get_h()
{
	int i,k=0;
	for (i=1;i<=len;i++)
	{
		int j=sa[rk[i]-1];
		if (k>0) k--;
		while (s[i+k]==s[j+k]) k++;
		h[rk[i]]=k;
	}
}
void get_sa()
{
	int i,k,p,ln,m=130;
	for (i=1;i<=len;i++) rk[i]=s[i];
	for (i=0;i<=m;i++) Rsort[i]=0;
	for (i=1;i<=len;i++) Rsort[rk[i]]++;
	for (i=1;i<=m;i++) Rsort[i]+=Rsort[i-1];
	for (i=len;i>=1;i--) sa[Rsort[rk[i]]--]=i;
	p=0;ln=1;
	while (p<len)
	{
		for (k=0,i=len-ln+1;i<=len;i++) y[++k]=i;
		for (i=1;i<=len;i++) if (sa[i]-ln>0) y[++k]=sa[i]-ln;
		for (i=1;i<=len;i++) wr[i]=rk[y[i]];
		for (i=0;i<=m;i++) Rsort[i]=0;
		for (i=1;i<=len;i++) Rsort[wr[i]]++;
		for (i=1;i<=m;i++) Rsort[i]+=Rsort[i-1];
		for (i=len;i>=1;i--) sa[Rsort[wr[i]]--]=y[i];
		memcpy(wr,rk,sizeof(wr));
		p=1;rk[sa[1]]=1;
		for (i=2;i<=len;i++)
		{
			if (!cmp(sa[i],sa[i-1],ln)) p++;
			rk[sa[i]]=p;
		}m=p;ln*=2;
	}sa[0]=s[0]=0;
}
int main()
{
	int T,i;scanf("%d
",&T);
	while (T--)
	{
		scanf("%s",s+1);
		len=strlen(s+1);
		get_sa();get_h();
		LL ans=0;h[1]=0;
		for (i=1;i<=len;i++)
		{
			LL ls=sa[i];
			ans+=len-ls+1-h[i];
		}printf("%lld
",ans);
	}
	return 0;
}