hdu 5008 查找字典序第k小的子串

Boring String Problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1848    Accepted Submission(s): 492

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

 

/*
hdu 5008 查找字典序第k小的子串

problem:
给你一个字符串，每次寻找字典序第k小的子串(不同的)的开始,结尾坐标。

solve:
首先可以知道height本身就是按字典序来弄的,对于sa[i-1]和sa[i]而言只要减去
它们的公共部分即height[i],就是不同的子串.
可以借此处理出来[1,n]即到字典序第i大的后缀时总共有多少个不同的子串.然后
利用二分查找就能找到位置
还要注意就是当有多个的时候输出最小的位置,可以判断当前子串是否是公共前缀的
一部分.往后枚举找出最小的位置即可
a
aa
aaa
当你要找最小的串a时可能会找到 3 3.因为它也是aa,aaa的一部分,所以需要枚举一下

hhh-2016-08-12 19:24:38
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <map>
#include <set>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
typedef unsigned int ul;
const int INF = 0x3f3f3f3f;
const int  maxn = 100000+10;
const int mod = 1e9+7;

int t1[maxn],t2[maxn],c[maxn];
bool cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b] &&r[l+a] == r[l+b];
}

void get_sa(int str[],int sa[],int Rank[],int height[],int n,int m)
{
    n++;
    int p,*x=t1,*y=t2;
    for(int i = 0; i < m; i++) c[i] = 0;
    for(int i = 0; i < n; i++) c[x[i] = str[i]]++;
    for(int i = 1; i < m; i++) c[i] += c[i-1];
    for(int i = n-1; i>=0; i--) sa[--c[x[i]]] = i;
    for(int j = 1; j <= n; j <<= 1)
    {
        p = 0;
        for(int i = n-j; i < n; i++) y[p++] = i;
        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
        for(int i = 0; i < m; i++) c[i] = 0;
        for(int i = 0; i < n; i++) c[x[y[i]]]++ ;
        for(int i = 1; i < m; i++) c[i] += c[i-1];
        for(int i = n-1; i >= 0; i--)  sa[--c[x[y[i]]]] = y[i];

        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for(int i = 1; i < n; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)? p-1:p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    n--;
    for(int i = 0; i <= n; i++)
        Rank[sa[i]] = i;
    for(int i = 0; i < n; i++)
    {
        if(k) k--;
        int j = sa[Rank[i]-1];
        while(str[i+k] == str[j+k]) k++;
        height[Rank[i]] = k;
    }

}

int Rank[maxn],height[maxn];
int sa[maxn];
char str[maxn];
int r[maxn];
ll num[maxn];
int lpos,rpos;
int len ;


void fin(ll x)
{
    int l =1 ,r = len;
    int cur = 0;
    while(l <= r)
    {
        int mid = (l+r) >> 1;
        if(num[mid] >= x)
        {
            cur = mid;
            r = mid-1;
        }
        else
            l = mid+1;
    }
//    cout << "cur:" <<cur <<endl;
    x = x - num[cur-1];
    lpos = sa[cur]+1;
    rpos = sa[cur]+height[cur]+x;
    int tlen = rpos-lpos+1;
    if(cur+1 <= len && tlen <= height[cur+1])
    {
        for(int i = cur + 1; i <= len; i++)
        {
            if(height[i] >= tlen)
            {
                if(lpos > sa[i]+1)
                {
                    lpos = sa[i]+1;
                    rpos = lpos+tlen-1;
                }
            }
            else
                break;
        }
    }
    return ;
}

int main()
{
    while(scanf("%s",str) != EOF)
    {
        len = strlen(str);
        for(int i = 0; i <len; i ++)
        {
            r[i] = str[i];
        }
        r[len] = 0;
        num[0] = 0;
        get_sa(r,sa,Rank,height,len,200);
        for(int i = 1; i <= len; i++)
        {
            num[i] = (ll)(len - sa[i] - height[i]);
            num[i] += num[i-1];
        }
        int n;
        ll x;
        scanf("%d",&n);
        lpos = 0,rpos = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%I64d",&x);
            ll k = (ll)(lpos^rpos^x)+1LL;
//            cout << "k:" << k <<endl;
            lpos = rpos = 0;
            if(k > num[len])
            {
                printf("0 0\n");
                lpos = 0,rpos = 0;
                continue;
            }
            fin(k);
            printf("%d %d\n",lpos,rpos);
        }
    }
    return 0;
}