hdu 5008(2014 ACM/ICPC Asia Regional Xi'an Online ) Boring String Problem(尾碼數組&二分)，hduicpc

hdu 5008(2014 ACM/ICPC Asia Regional Xi'an Online ) Boring String Problem(尾碼數組&二分)，hduicpc
Boring String ProblemTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 219    Accepted Submission(s): 45


Problem DescriptionIn 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 si...j of the string s = a1a2 ...an(1 ≤ i ≤ j ≤ n) is the string aiai+1 ...aj. Two substrings sx...y and sz...w are cosidered to be distinct if sx...y ≠ Sz...w
 
InputThe 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)
 
OutputFor 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 sl...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 s1...n is the whole string)
 
Sample Input

aaa40235

 
Sample Output

1 11 31 20 0

 
Source2014 ACM/ICPC Asia Regional Xi'an Online 
Recommendhujie   |   We have carefully selected several similar problems for you:  5017 5016 5015 5014 5013  題意：給你一個長度不超過1e5的字串。把它的所有子串去重後排序。然後要你輸出第k大的字串的位置l,r。如果有多個位置輸出l最小的。思路：比賽時看到這題感覺和SPOJ-7258 Lexicographical Substring Search很像。不過那是學尾碼自動機時看到了。由於尾碼自動機是在是不是很好懂。。。所以最後還是放棄了。但是網上有這題的題解。但是很那題有點差別的是這題要求輸出字串的位置。於是就不知道怎麼搞了。於是就用相對熟悉的尾碼數組想了下。畢竟這題感覺n*log(n)是可過的。然後就開幹了。我們先算出來每個尾碼sa[i]比sa[i-1]多出多少個不同的子串。明顯為len-sa[i]-height[i]存到val[i]。而sa[i]就對應len-sa[i]-height[i]個子串了。然後用val[i]構造線段樹。然後找排名第k的字串怎麼找呢。就類似二分的思想了。如果線段樹左子樹字元大於等於k個就到左子樹。不行就到右子樹找。這樣就可以找到第k大串對應的sa[i]和第k串的長度len。寫完這裡就卡了下。因為怎麼處理l最小的問題呢。不可能在i附近找和sa[i]lcp>=len且sa[i]最小的吧。想了下最發雜的就是全a的情況。這樣就退化成O(n^2)了。那sa就白寫了。冷靜了下。一想。就出來了。我們可以二分求出最左邊和sa[i]lcp>=len的位置left。在二分出sa[i]最右邊lcp>=len的位置right。然後答案就是left->right中sa最小的。這個可以用rmq維護。然後接1A了。不過有人就是按我先前的前後直接找的。居然過了。可見資料還是蠻水的。詳細見代碼：

#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;}