HDU - 5008 Boring String Problem （后缀数组+二分+RMQ）

Boring String Problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2160    Accepted Submission(s): 588


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 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

 

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 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

aaa
4
0
2
3
5

 

Sample Output

1 1
1 3
1 2
0 0

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

Recommend

hujie

 

题意：给你你一个字符串，求第k小的子串，，有很多询问

解题思路：后缀数组！第一次自己思考出来了。首先先统计不同的后缀对子串的数量的贡献，然后求个前缀和，通过二分可以快速找到第k小的子串！能想到这里这题就差不多了。但是这样找到的子串的下标是最大的，我们要求最小！所以我们要从L往后找最小的SA，这里通过一个RMQ维护height数组，通过二分R，可以快速定位可能有最小的下标的SA区间，，然后再通过一个RMQ查询这个区间的最小下标即可！但是这题暴力往后扫可过……证明没有100000个a这样的数据。这里给出暴力代码

#include <iostream>
#include <deque>
#include <stdio.h>
#include <map>
#include <string.h>
#include <algorithm>
#include <vector>
#include <math.h>
#include <stack>
#include <queue>
#include <set>
using namespace std;
inline void scan_d(int &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}

inline void scan_d(long long int &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}
typedef long long int ll;

const int MAXN = 100005; //用DC3要开三倍大小

int wa[MAXN], wb[MAXN], wv[MAXN], JS[MAXN]; //计算SA用的辅助数组

int rk[MAXN], height[MAXN], SA[MAXN]; //三个常用数组

/***后缀数组倍增解法***/
int cmp(int *r, int a, int b, int l)
{
return r[a] == r[b] && r[a + l] == r[b + l];
}

void DA(int *r, int *SA, int n, int m)
{
int i, j, p, *x = wa, *y = wb, *t;
for (i = 0; i < m; i++)
JS[i] = 0;
for (i = 0; i < n; i++)
JS[x[i] = r[i]]++;
for (i = 1; i < m; i++)
JS[i] += JS[i - 1];
for (i = n - 1; i >= 0; i--)
SA[--JS[x[i]]] = i;
for (j = 1, p = 1; p < n; j *= 2, 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 < n; i++)
wv[i] = x[y[i]];
for (i = 0; i < m; i++)
JS[i] = 0;
for (i = 0; i < n; i++)
JS[wv[i]]++;
for (i = 1; i < m; i++)
JS[i] += JS[i - 1];
for (i = n - 1; i >= 0; i--)
SA[--JS[wv[i]]] = y[i];
for (t = x, x = y, y = t, 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++;
}
return;
}
/*******************/

/***后缀数组DC3解法***/
#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))
#define G(x) ((x) < tb ? (x)*3 + 1 : ((x)-tb) * 3 + 2)
int c0(int *r, int a, int b)
{
return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}
int c12(int k, int *r, int a, int b)
{
if (k == 2)
return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1);
else
return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];
}
void sort(int *r, int *a, int *b, int n, int m)
{
int i;
for (i = 0; i < n; i++)
wv[i] = r[a[i]];
for (i = 0; i < m; i++)
JS[i] = 0;
for (i = 0; i < n; i++)
JS[wv[i]]++;
for (i = 1; i < m; i++)
JS[i] += JS[i - 1];
for (i = n - 1; i >= 0; i--)
b[--JS[wv[i]]] = a[i];
return;
}
void DC3(int *r, int *SA, int n, int m)
{
int i, j, *rn = r + n, *SAn = SA + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;
r
= r[n + 1] = 0;
for (i = 0; i < n; i++)
if (i % 3 != 0)
wa[tbc++] = i;
sort(r + 2, wa, wb, tbc, m);
sort(r + 1, wb, wa, tbc, m);
sort(r, wa, wb, tbc, m);
for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;
if (p < tbc)
DC3(rn, SAn, tbc, p);
else
for (i = 0; i < tbc; i++)
SAn[rn[i]] = i;
for (i = 0; i < tbc; i++)
if (SAn[i] < tb)
wb[ta++] = SAn[i] * 3;
if (n % 3 == 1)
wb[ta++] = n - 1;
sort(r, wb, wa, ta, m);
for (i = 0; i < tbc; i++)
wv[wb[i] = G(SAn[i])] = i;
for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
SA[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
for (; i < ta; p++)
SA[p] = wa[i++];
for (; j < tbc; p++)
SA[p] = wb[j++];
return;
}
/***********************/

//计算rank和height数组
void calheight(int *r, int *SA, int n)
{
// memset(height,0,sizeof(height));
// memset(rk,0,sizeof(rk));
int i, j, k = 0;
for (i = 1; i <= n; i++)
rk[SA[i]] = i;
for (i = 0; i < n; height[rk[i++]] = k)
for (k ? k-- : 0, j = SA[rk[i] - 1]; r[i + k] == r[j + k]; k++)
;
}
int N; //字符串长度
/*****RMQ部分*****/
int ST[MAXN][16];
void RMQ(int *a)
{
int m = (int)(log(N * 1.0) / log(2.0));
for (int i = 1; i <= N; i++)
ST[i][0] = a[i];
for (int j = 1; j <= m; j++)
for (int i = 1; i + (1 << j) - 1 <= N; i++)
ST[i][j] = min(ST[i][j - 1], ST[i + (1 << (j - 1))][j - 1]);
}
int query(int a, int b)
{
int k = int(log(b - a + 1.0) / log(2.0));
return min(ST[a][k], ST[b - (1 << k) + 1][k]);
}

char sstr[MAXN]; //主字符串
int str[MAXN];
ll subnum[MAXN];

bool judge(int m,ll k){
if(subnum[m]<k)
return true;
else
return false;
}

int main()
{

while (~scanf("%s", sstr))
{
N=strlen(sstr);
for (int i = 0; i < N ; i++)
str[i] = sstr[i] - 'a' + 1;

str
= 0; //必须要末尾补0！！！！！

DA(str, SA, N + 1, 28);//N是没有补0的大小，算SA时要把末尾0计算进去，所以要N+1
calheight(str, SA, N);//计算height时不用末尾0
RMQ(height);

//cout<<query(2,5)<<endl;
//调试
// for(int i=0;i<=N+1;i++)
// cout<<SA[i]<<" ";
// cout<<endl;
// for(int i=0;i<=N+1;i++)
// cout<<rk[i]<<" ";
// cout<<endl;
// for(int i=0;i<=N+1;i++)
// cout<<height[i]<<" ";
// cout<<endl;

subnum[0]=0;
for(int i=1;i<=N;i++){
subnum[i]=subnum[i-1]+N-SA[i]-height[i];
// cout<<subnum[i]<<" ";
}
int q;
scan_d(q);
ll v,k;
ll l=0,r=0;
int m;
for(int i=0;i<q;i++){
scan_d(v);

k=(l^r^v)+1;

if(k>subnum
){
printf("%d %d\n",0,0);
l=0,r=0;
continue;
}

// k=v;
l=1;
r=N;
while(l<r){
m=(l+r)/2;
if(judge(m,k)){
l=m+1;
}
else
r=m;
}

int len=k-subnum[l-1]+height[l];
// cout<<l<<endl;
int L=SA[l]+1;
int R=L+len-1;
// cout<<L<<" "<<R<<endl;

int ii;
int lll=l;
for(ii=l+1;ii<=N;ii++){
if(height[ii]<len)
break;
else{
if(SA[ii]<SA[lll])
lll=ii;
}
}

l=SA[lll]+1;
r=l+len-1;

printf("%I64d %I64d\n",l,r);
}

}

return 0;
}