hdu 5008 Boring String Problem

Boring String Problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 291    Accepted Submission(s): 71

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

 

一道很好的后缀数组题目，比赛时意识到了这题会用后缀数组，但是一时有没想到如何查找第k大下标最靠左的位置。比赛完后，才知道可以二分查找。

这道题目求有多少不同的子串，在我的blog里有原题，以及怎么height数组为什么可以二分查找，我之前做过的题也有解释，可以参考我的后缀数组这块。虽然刷过不少的后缀数组题，但是真正的比赛还是写不出来啊，看来水平远远不够。注意一下，网上的题解有些没有二分查找下标最靠左的，这样其实会超时的，测试数据可能水了些吧。

比如全a的话，如果要查排名第1的位置，那么不得查到最后一个，这样显然超时，测试数据应该给出这样的例子，这题还要注意long long，被坑了好多次。

代码：

#include<cstdio>#include<iostream>#include<cstring>#define Maxn 100010#define ll long longusing namespace std;char s[Maxn];int r[Maxn],sa[Maxn],rk[Maxn],height[Maxn];int wa[Maxn],wb[Maxn],rs[Maxn],wv[Maxn];ll a,b,k,v;int cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(int n,int m){    int i,j,p,*x=wa,*y=wb;    for(i=0;i<m;i++) rs[i]=0;    for(i=0;i<n;i++) rs[x[i]=r[i]]++;    for(i=1;i<m;i++) rs[i]+=rs[i-1];    for(i=n-1;i>=0;i--) sa[--rs[x[i]]]=i;    for(j=1,p=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<m;i++) rs[i]=0;        for(i=0;i<n;i++) rs[wv[i]=x[y[i]]]++;        for(i=1;i<m;i++) rs[i]+=rs[i-1];        for(i=n-1;i>=0;i--) sa[--rs[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++;    }}void calheight(int n){    int i,j,k=0;    for(int i=1;i<n;i++) rk[sa[i]]=i;    for(int i=1;i<n;height[rk[i++]]=k){        if(k) k--;        for(j=sa[rk[i]-1];r[i+k]==r[j+k];k++);    }}int p[Maxn];ll d[Maxn][20],d1[Maxn][20],sum[Maxn];void rmq_init(int n,int *height,ll (*d)[20]){    p[0]=-1;    for(int i=1;i<=n;i++)        p[i]=i&i-1?p[i-1]:p[i-1]+1;    for(int i=1;i<=n;i++) d[i][0]=height[i];    for(int j=1;j<=p[n];j++)        for(int i=1;i+(1<<j)-1<=n;i++)            d[i][j]=min(d[i][j-1],d[i+(1<<j-1)][j-1]);}ll rmq_ask(int l,int r,ll (*d)[20]){    int k=p[r-l+1];    return min(d[l][k],d[r-(1<<k)+1][k]);}ll lcp(int a,int b){    if(a<b) swap(a,b);    return rmq_ask(b+1,a,d);}int b_search1(int l,int r,ll k){    while(l<r){        int mid=l+r+1>>1;        if(sum[mid]<k) l=mid;        else r=mid-1;    }    return l;}ll b_search2(int s,int l,int r,ll k){    while(l<r){        int mid=l+r+1>>1;        if(lcp(s,mid)<k) r=mid-1;        else l=mid;    }    if(lcp(s,l)>=k) return rmq_ask(s,l,d1);    return a;}int main(){    int i,t,n;    while(~scanf("%s",s+1)){        for(i=1;s[i];i++)            r[i]=s[i]-'a'+1;        r[0]=r[n=i]=sum[0]=0;        da(n,27);        calheight(n);        for(i=1;i<n;i++) sum[i]=n-sa[i]-height[i];        for(i=1;i<n;i++) sum[i]+=sum[i-1];        rmq_init(n,height,d);        rmq_init(n,sa,d1);        a=b=0;        scanf("%d",&t);        while(t--){            scanf("%I64d",&v);            k=(a^b^v)+1;            if(k<=sum[n-1]){                int id=b_search1(0,n-1,k)+1;                a=sa[id];                a=min(a,b_search2(id,id+1,n-1,k-sum[id-1]+height[id]));                b=a+k-sum[id-1]+height[id]-1;            }            else a=b=0;            printf("%I64d %I64d\n",a,b);        }    }return 0;}