hdu5008 Boring String Problem(后缀数组)

Boring String Problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 927    Accepted Submission(s): 263

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

aaa40235

 

Sample Output

1 11 31 20 0

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

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解题思路：

后缀数组lcp求好，直接暴力找就过了。。。

#include <iostream>#include <cstdio>#include <cstring>#define ll long long#define maxn 100010using namespace std;char s[maxn];int n,k;int rank[maxn],sa[maxn],tmp[maxn],lcp[maxn];bool cmp(int x,int y){    if(rank[x]!=rank[y]) return rank[x]<rank[y];    int sx=x+k<=n ? rank[x+k]:-1;    int sy=y+k<=n ? rank[y+k]:-1;    return sx<sy;}void build_sa(){    n=strlen(s);    for(int i=0;i<=n;i++){        sa[i]=i;        rank[i]=i<n ? s[i]:-1;    }    for(k=1;k<=n;k<<=1){        sort(sa,sa+n+1,cmp);        tmp[sa[0]]=0;        for(int i=1;i<=n;i++){            tmp[sa[i]]=tmp[sa[i-1]]+(cmp(sa[i-1],sa[i]) ? 1:0);        }        for(int i=0;i<=n;i++) rank[i]=tmp[i];    }}void build_lcp(){    n=strlen(s);    for(int i=0;i<=n;i++) rank[sa[i]]=i;    int h=0;    lcp[0]=0;    for(int i=0;i<n;i++){        int j=sa[rank[i]-1];        if(h>0) h--;        for(;j+h<n&&i+h<n;h++){            if(s[j+h]!=s[i+h]) break;        }        lcp[rank[i]-1]=h;    }}ll sum[maxn];int next[maxn];void solve(){    sum[0]=0;    for(int i=1;i<=n;i++){        sum[i]=sum[i-1]+n-sa[i]-lcp[i-1];    }    int q;    scanf("%d",&q);    int l=0,r=0;    ll v;    memset(next,-1,sizeof next);    for(int i=0;i<q;i++){        scanf("%I64d",&v);        v=(l^r^v)+1;        if(v>sum[n]){            l=0,r=0;            printf("0 0\n");            continue;        }        int p=lower_bound(sum+1,sum+n+1,v)-sum;        r=v-sum[p-1]+lcp[p-1];        l=sa[p];        for(int j=p;j<n;j++){            if(lcp[j]<r) break;            if(sa[j+1]<l) l=sa[j+1];        }        r=l+r-1;        l++,r++;        printf("%d %d\n",l,r);    }}int main(){    while(~scanf("%s",s)){        build_sa();        build_lcp();        solve();    }    return 0;}