hdu 5008(2014 ACM/ICPC Asia Regional Xi'an Online ) Boring String Problem(后缀数组&二分)

Boring String Problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 219    Accepted Submission(s): 45

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 = a ₁a ₂ ...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| ≤ 10 ⁵) with only lowercase letters.

Next line contains a postive integer q(1 ≤ q ≤ 10 ⁵), 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 < 2 ⁶³, “⊕” 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

 

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  题意：

给你一个长度不超过1e5的字符串。把它的所有子串去重后排序。然后要你输出第k大的字符串的位置l,r。如果有多个位置输出l最小的。

思路：

比赛时看到这题感觉和SPOJ-7258 Lexicographical Substring Search很像。

不过那是学后缀自动机时看到了。由于后缀自动机是在是不是很好懂。。。所以最后还是放弃了。但是网上有这题的题解。但是很那题有点差别的是这题要求输出字符串的位置。于是就不知道怎么搞了。于是就用相对熟悉的后缀数组想了下。毕竟这题感觉n*log(n)是可过的。然后就开干了。我们先算出来每个后缀sa[i]比sa[i-1]多出多少个不同的子串。明显为len-sa[i]-height[i]存到val[i]。而sa[i]就对应len-sa[i]-height[i]个子串了。然后用val[i]构造线段树。然后找排名第k的字符串怎么找呢。就类似二分的思想了。如果线段树左子树字符大于等于k个就到左子树。不行就到右子树找。这样就可以找到第k大串对应的sa[i]和第k串的长度len。写完这里就卡了下。因为怎么处理l最小的问题呢。不可能在i附近找和sa[i]lcp>=len且sa[i]最小的吧。想了下最发杂的就是全a的情况。这样就退化成O(n^2)了。那sa就白写了。

冷静了下。一想。就出来了。我们可以二分求出最左边和sa[i]lcp>=len的位置left。在二分出sa[i]最右边lcp>=len的位置right。然后答案就是left->right中sa最小的。这个可以用rmq维护。然后接1A了。不过有人就是按我先前的前后直接找的。居然过了。可见数据还是蛮水的。

详细见代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
typedef long long ll;
#define lson L,mid,ls
#define rson mid+1,R,rs
char txt[maxn];
int sa[maxn],T1[maxn],T2[maxn],ct[maxn],he[maxn],rk[maxn],n,m,le,ri;
int rmq[25][maxn],lg[maxn],id[25][maxn],pos,len;
ll num[maxn<<2];
void build(int L,int R,int rt)
{
    if(L==R)
    {
        num[rt]=n-sa[L]-he[L];
        return;
    }
    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;
    build(lson);
    build(rson);
    num[rt]=num[ls]+num[rs];
}
void getsa(char *st)
{
    int i,k,p,*x=T1,*y=T2;
    for(i=0; i<m; i++) ct[i]=0;
    for(i=0; i<n; i++) ct[x[i]=st[i]]++;
    for(i=1; i<m; i++) ct[i]+=ct[i-1];
    for(i=n-1; i>=0; i--)
        sa[--ct[x[i]]]=i;
    for(k=1,p=1; p<n; k<<=1,m=p)
    {
        for(p=0,i=n-k; i<n; i++) y[p++]=i;
        for(i=0; i<n; i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(i=0; i<m; i++) ct[i]=0;
        for(i=0; i<n; i++) ct[x[y[i]]]++;
        for(i=1; i<m; i++) ct[i]+=ct[i-1];
        for(i=n-1; i>=0; i--) sa[--ct[x[y[i]]]]=y[i];
        for(swap(x,y),p=1,x[sa[0]]=0,i=1; i<n; i++)
            x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
    }
}
void gethe(char *st)
{
    int i,j,k=0;
    for(i=0;i<n;i++) rk[sa[i]]=i;
    for(i=0;i<n-1;i++)
    {
        if(k) k--;
        j=sa[rk[i]-1];
        while(st[i+k]==st[j+k]) k++;
        he[rk[i]]=k;
    }
}
void rmq_init()
{
    int i,j;
    for(i=0;i<n;i++)
    {
        rmq[0][i]=he[i];
        id[0][i]=sa[i];
    }
    for(i=1;i<=lg[n];i++)
        for(j=0;j+(1<<i)-1<n;j++)
        {
            rmq[i][j]=min(rmq[i-1][j],rmq[i-1][j+(1<<(i-1))]);
            id[i][j]=min(id[i-1][j],id[i-1][j+(1<<(i-1))]);
        }
}
int rmq_min(int l,int r)
{
    if(l>r)
        return 0;
    int tmp=lg[r-l+1];
    return min(rmq[tmp][l],rmq[tmp][r-(1<<tmp)+1]);
}
int rmq_id(int l,int r)
{
    int tmp=lg[r-l+1];
    return min(id[tmp][l],id[tmp][r-(1<<tmp)+1]);
}
void prermq()
{
    int  i;
    lg[0]=-1;
    for(i=1;i<maxn;i++)
        lg[i]=lg[i>>1]+1;
}
void qu(int L,int R,int rt,ll k)
{
    if(k>num[rt])
    {
        pos=-1;
        le=ri=0;
        return;
    }
    if(L==R)
    {
        pos=L;
        len=he[L]+k;
        return;
    }
    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;
    if(num[ls]>=k)
        qu(lson,k);
    else
        qu(rson,k-num[ls]);
}
int binl(int x)
{
    int low=1,hi=x-1,mid,ans=x;

    while(low<=hi)
    {
        mid=(low+hi)>>1;
        if(rmq_min(mid+1,x)>=len)
            ans=mid,hi=mid-1;
        else
            low=mid+1;
    }
    return ans;
}
int binr(int x)
{
    int low=x+1,hi=n,mid,ans=x;

    while(low<=hi)
    {
        mid=(low+hi)>>1;
        if(rmq_min(x+1,mid)>=len)
            ans=mid,low=mid+1;
        else
            hi=mid-1;
    }
    return ans;
}
inline ll ReadInt()
{
    char ch = getchar();
    if (ch==EOF) return -1;
    ll data = 0;
    while (ch < '0' || ch > '9')
    {
        ch = getchar();
        if (ch==EOF) return -1;
    }
    do
    {
        data = data*10 + ch-'0';
        ch = getchar();
    } while (ch >= '0' && ch <= '9');
    return data;
}

inline void putit(int x)
{
    if (x/10>0) putit(x/10);
    putchar(x%10+'0');
}
int main()
{
    int q,lll,rrr;
    ll kth;

    prermq();
    while(~scanf("%s",txt))
    {
        m=150,n=strlen(txt);
        n++;
        getsa(txt);
        gethe(txt);
        rmq_init();
        n--;
        build(1,n,1);
        scanf("%d",&q);
        le=ri=0;
        while(q--)
        {
            kth=ReadInt();
            kth=(le^ri^kth)+1;
            qu(1,n,1,kth);
            if(pos==-1)
                putit(le),putchar(' '),putit(ri),putchar('\n');
            else
            {
                lll=binl(pos);
                rrr=binr(pos);
                le=rmq_id(lll,rrr)+1;
                ri=le+len-1;
                putit(le),putchar(' '),putit(ri),putchar('\n');
            }
        }
    }
    return 0;
}