hdu3518之后缀数组详解

Boring counting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1320 Accepted Submission(s): 532



Problem Description

035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.

Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and
[1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).



Input

The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).



Output

For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.



Sample Input

aaaa
ababcabb
aaaaaa
#

Sample Output

2
3
3
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;

const int MAX=1000+10;
int *rank,r[MAX],sa[MAX],height[MAX];
int wa[MAX],wb[MAX],wn[MAX];
char s[MAX];

bool cmp(int *r,int a,int b,int l){
	return r[a] == r[b] && r[a+l] == r[b+l];//r[a]表示第一关键字,r[a+l]表示第二关键字 
}

void makesa(int *r,int *sa,int n,int m){
	int *x=wa,*y=wb,*t;//x记录的是排序完的结果,相当于rank,y记录第二关键字
	 
	//第一次基数排序
	for(int i=0;i<m;++i)wn[i]=0;//对所有的子串个数清0 
	for(int i=0;i<n;++i)wn[x[i]=r[i]]++;//子串r[i]的个数(这里的子串长度为1即子串是r[i]) 
	for(int i=1;i<m;++i)wn[i]+=wn[i-1];//统计<=子串i的个数(子串长度为1,i表示的是子串的ASCII) 
	//wn[x[i]]表示<=x[i]的子串有多少个(排名从0开始,所以--wn[]),i=n-1开始是保证相同的子串位置在后面的就排在后面
	for(int i=n-1;i>=0;--i)sa[--wn[x[i]]]=i;//排名为--wn[x[i]]的子串是suffix(i) 
	
	//倍增进行基数排序直到全部子串的排名不同
	for(int i=0,j=1,p=1;p<n;j=j*2,m=p){//j表示要合并的子串的长度,m=p表示需要排序的最大值为p即排名 
		for(p=0,i=n-j;i<n;++i)y[p++]=i;//对第二关键字数组越界的子串先进行第二关键字排序
		for(i=0;i<n;++i)if(sa[i]>=j)y[p++]=sa[i]-j;//对剩余的子串进行第二关键字排序
		//通过第二关键字和第一关键字进行基数排序
		for(i=0;i<m;++i)wn[i]=0;
		for(i=0;i<n;++i)wn[x[y[i]]]++;//x记录的是上次排序完的结果(排名,也是第一关键字),相当于rank
		for(i=1;i<m;++i)wn[i]+=wn[i-1];
		//第二关键字(排名)从大到小,在第二关键字为i的那个子串y[i]的第一关键字为x[y[i]],第一关键字x[y[i]]前面有wn[]个则排名--wn[](排名从0开始) 
		for(i=n-1;i>=0;--i)sa[--wn[x[y[i]]]]=y[i];//计算排名为--wn[]的是子串y[i]
		//计算新的x相等的子串排名得相同,因为上一次的排名x要用到所以先用y保存(交换x,y即可) 
		for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i){
			x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;//如果子串的第一关键字和第二关键字相等则排名一样即p-1 
		} 
	}
	rank=x;//排序完后r已经是完全是rank了 
}

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

int main(){
	while(~scanf("%s",s),s[0] != '#'){
		int len=0;
		for(len=0;s[len] != '\0';++len)r[len]=s[len];//转化成整数
		r[len]=0;
		makesa(r,sa,len+1,256);//len+1表示在字符串后面+0确保子串比较不会越界(即越界前一定能比出大小)
		calheight(r,sa,len+1);
		int L=INF,R=-INF,sum=0;
		for(int i=1;i<=len/2;++i){//长度为i公共子串
			L=INF,R=-INF;
			for(int j=2;j<=len;++j){//height[1]=0,所以j可以从2开始 
				if(height[j]>=i){
					L=min(L,sa[j]);
					L=min(L,sa[j-1]);
					R=max(R,sa[j]);
					R=max(R,sa[j-1]);
				}else{
					if(L+i<=R)++sum;
					L=INF,R=-INF;
				}
			}
			if(L+i<=R)++sum;
		}
		cout<<sum<<endl;
	}
	return 0;
}