Hdu5008-Boring String Problem(后缀数组)

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

 

题意：找到字符串中第k小的字串，并且输出下标最小的那一个。

 

解析：用后缀数组处理，每个后缀不同的子串个数是len-sa[i]-h[i],所以可以二分去找，但是找到了并不一定是下标最小的，所以还要继续往后面更新。

 

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef __int64 LL;
typedef pair<int,int> par;
const int maxn=100005;
char S[maxn];
int A[maxn];
LL sum[maxn];
int wa[maxn],wb[maxn],wv[maxn],WS[maxn],sa[maxn];
bool cmp(int *r,int a,int b,int l){ return r[a]==r[b]&&r[a+l]==r[b+l]; }
void DA(int *r,int n,int m)  //模板
{
    int i,j,p;
    int *x=wa,*y=wb;
    for(i=0;i<m;i++) WS[i]=0;
    for(i=0;i<n;i++) WS[x[i]=r[i]]++;
    for(i=1;i<m;i++) WS[i]+=WS[i-1];
    for(i=n-1;i>=0;i--) sa[--WS[x[i]]]=i;

    for(p=1,j=1;p<n;j<<=1,m=p)
    {
        for(p=0,i=n-j;i<n;i++) y[p++]=i;
        for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;
        for(i=0;i<n;i++) wv[i]=x[y[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--) sa[--WS[wv[i]]]=y[i];
        swap(x,y);
        for(p=1,x[sa[0]]=0,i=1;i<n;i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
}
int rk[maxn],h[maxn];
void GetHeight(int *r,int n)
{
    for(int i=0;i<=n;i++) rk[sa[i]]=i;
    int k=0;
    h[0]=0;
    for(int i=0;i<n;i++)
    {
        if(k) k--;  //先减1
        int j=sa[rk[i]-1];//排名在前面的
        while(r[i+k]==r[j+k]) k++; //相同一直加
        h[rk[i]]=k;
    }
}
int BIS(int x,int y,LL k)//找到第一个大于k的位置
{
    while(x<=y)
    {
        int mid=(x+y)/2;
        if(sum[mid]>=k) y=mid-1;
        else x=mid+1;
    }
    return y+1;
}
int main()
{
    while(scanf("%s",S)!=EOF)
    {
        int len=strlen(S);
        for(int i=0;i<len;i++) A[i]=S[i];
        A[len]=0;
        DA(A,len+1,130); //后缀数组处理
        GetHeight(A,len); //得到lcp值
        sum[0]=0;
        for(int i=1;i<=len;i++)
        {
            int a=sa[i];
            int b=h[i];
            sum[i]=sum[i-1]+len-a-b; //计算不同子串个数
        }
        int Q;
        LL l=0,r=0,v;
        scanf("%d",&Q);
        while(Q--)
        {
            scanf("%lld",&v);
            LL k=(l^r^v)+1;
            if(k>sum[len]) printf("%lld %lld
",l=0,r=0); //无解的情况
            else
            {
                int p=BIS(1,len,k); //二分找
                l=sa[p];
                int sl=h[p]+k-sum[p-1]; //长度
                int cur=p;
                while(++cur<=len&&h[cur]>=sl) l=min(l,(LL)sa[cur]);//往后面找位置最小的，不知道这种方式为何不会超时
                r=l+sl-1;
                l++; r++;
                printf("%lld %lld
",l,r);
            }
        }
    }
    return 0;
}