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 = 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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 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,rschar 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;}