HDOJ 5008 Boring String Problem

后缀数组+RMQ+二分

后缀数组二分确定第K不同子串的位置 , 二分LCP确定可选的区间范围 , RMQ求范围内最小的sa

Boring String Problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 661    Accepted Submission(s): 183

Problem Description

In this problem, you are given a string s and q queries.

For each query, you should answer that when all distinct substrings of string s were sorted lexicographically, which one is the k-th smallest. 

A substring s_i...j of the string s = a₁a₂ ...a_n(1 ≤ i ≤ j ≤ n) is the string a_ia_i+1 ...a_j. Two substrings s_x...y and s_z...w are cosidered to be distinct if s_x...y ≠ S_z...w

 

Input

The input consists of multiple test cases.Please process till EOF. 

Each test case begins with a line containing a string s(|s| ≤ 10⁵) with only lowercase letters.

Next line contains a postive integer q(1 ≤ q ≤ 10⁵), the number of questions.

q queries are given in the next q lines. Every line contains an integer v. You should calculate the k by k = (l⊕r⊕v)+1(l, r is the output of previous question, at the beginning of each case l = r = 0, 0 < k < 2⁶³, “⊕” denotes exclusive or)

 

Output

For each test case, output consists of q lines, the i-th line contains two integers l, r which is the answer to the i-th query. (The answer l,r satisfies that s_l...r is the k-th smallest and if there are several l,r available, ouput l,r which with the smallest l. If there is no l,r satisfied, output “0 0”. Note that s_1...n is the whole string)

 

Sample Input

aaa
4
0
2
3
5

 

Sample Output

1 1
1 3
1 2
0 0

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long int LL;

const int maxn=110100;
const int INF=0x3f3f3f3f;

int sa[maxn],rank[maxn],rank2[maxn],h[maxn],c[maxn],
	*x,*y,ans[maxn];
char str[maxn];

bool cmp(int* r,int a,int b,int l,int n)
{
	if(r[a]==r[b]&&a+l<n&&b+l<n&&r[a+l]==r[b+l])
		return true;
	return false;
}

void radix_sort(int n,int sz)
{
	for(int i=0;i<sz;i++) c[i]=0;
	for(int i=0;i<n;i++) c[x[y[i]]]++;
	for(int i=1;i<sz;i++) c[i]+=c[i-1];
	for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
}

void get_sa(char c[],int n,int sz=128)
{
	x=rank,y=rank2;
	for(int i=0;i<n;i++) x[i]=c[i],y[i]=i;
	radix_sort(n,sz);
	for(int len=1;len<n;len<<=1)
	{
		int yid=0;
		for(int i=n-len;i<n;i++) y[yid++]=i;
		for(int i=0;i<n;i++) if(sa[i]>=len) y[yid++]=sa[i]-len;

		radix_sort(n,sz);

		swap(x,y);
		x[sa[0]]=yid=0;

		for(int i=1;i<n;i++)
		{
			x[sa[i]]=cmp(y,sa[i],sa[i-1],len,n)?yid:++yid;
		}
		sz=yid+1;
		if(sz>=n) break;
	}
	for(int i=0;i<n;i++) rank[i]=x[i];
}

void get_h(char str[],int n)
{
	int k=0; h[0]=0;
	for(int i=0;i<n;i++)
	{
		if(rank[i]==0) continue;
		k=max(k-1,0);
		int j=sa[rank[i]-1];
		while(i+k<n&&j+k<n&&str[i+k]==str[j+k]) k++;
		h[rank[i]]=k;
	}
}

LL Range[maxn];

int bin(LL x,int n)
{
	int ans=-1;
	int low=0,high=n-1,mid;
	while(low<=high)
	{
		mid=(low+high)/2;
		if(Range[mid]<x)
		{
			ans=mid;
			low=mid+1;
		}
		else
		{
			high=mid-1;
		}
	}
	return ans;
}

int lcp[maxn][20],mmm[maxn][20];

void RMQ_init(int n)
{
	for(int i=0;i<n;i++)
	{
		lcp[i][0]=h[i];
		mmm[i][0]=sa[i];
	}
	lcp[0][0]=0x3f3f3f3f;
	int sz=floor(log(n*1.0)/log(2.0));
	for(int i=1;(1<<i)<=n;i++)
	{
		for(int j=0;j+(1<<i)-1<n;j++)
		{
			lcp[j][i]=min(lcp[j][i-1],lcp[j+(1<<(i-1))][i-1]);
			mmm[j][i]=min(mmm[j][i-1],mmm[j+(1<<(i-1))][i-1]);
		}
	}
}

int LCP(int l,int r,int n)
{
	if(l==r) return n-sa[l];
	l++;
	if(l>r) swap(l,r);
	int k=0;
	while(1<<(k+1)<=r-l+1) k++;
	return min(lcp[l][k],lcp[r-(1<<k)+1][k]);
}

int MMM(int l,int r)
{
	if(l>r) swap(l,r);
	int k=0;
	while(1<<(k+1)<=r-l+1) k++;
	return min(mmm[l][k],mmm[r-(1<<k)+1][k]);
}

int binID(int x,int n,int len)
{
	int ans=x;	
	int low=x,high=n-1,mid;
	while(low<=high)
	{
		mid=(low+high)/2;
		if(LCP(x,mid,n)>=len)
		{
			ans=mid;
			low=mid+1;
		}
		else high=mid-1;
	}
	return ans;
}
int main()
{
while(scanf("%s",str)!=EOF)
{
	int n=strlen(str);
	get_sa(str,n);
	get_h(str,n);
	RMQ_init(n);
	for(int i=0;i<n;i++)
	{
		Range[i]=(n-sa[i])-h[i];
		if(i-1>=0) Range[i]+=Range[i-1];
	}
	int q;
	scanf("%d",&q);
	int L=0,R=0;
	LL V;
	while(q--)
	{
		scanf("%I64d",&V);
		LL K=(L^R^V)+1LL;
		if(K>Range[n-1])
		{
			L=0;R=0;
			printf("%d %d
",L,R);
			continue;
		}
		int id=bin(K,n);
		LL jian=0;
		if(id>=0) jian=Range[id];
		LL res=K-jian; id++;
		int len=h[id]+res;	
		int hid=binID(id,n,len);
		int Left=MMM(id,hid);	
		printf("%d %d
",Left+1,Left+len);
		L=Left+1;R=Left+len;
	}
}
	return 0;
}