# Boring String Problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 255 Accepted Submission(s): 58

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

这题就是找出所有子串中的第k大。 而且要找出下标最小的。     这题相对而言简直弱爆了。   用后缀数组可以随便搞。   首先sa数组就已经按照字典序排好了。   对于每个sa 都会增加一定的子串。 增加的子串个数就是  n - sa$i$ - height$i$.   然后二分就可以找到第k大的位置。 然后为了找到下标最小的。 二分确定一个区间，在这个区间求sa数组的最小值。 搞个rmq进行最大最小值查询。   记住 后缀的前缀就是子串！


#include <stdio.h>

#include

#include

#include

#include

#include <string.h>

#include

#include

#include

#include <math.h>
using namespace std;
/ suffix array
倍增算法 O(nlogn)*待排序数组长度为n,放在0~n-1中，在最后面补一个0*\da(str ,sa,rank,height, n , );//注意是n;
例如： n = 8;
* num[] = { 1, 1, 2, 1, 1, 1, 1, 2, \$ }; 注意num最后一位为0，其他大于0
rank[] = { 4, 6, 8, 1, 2, 3, 5, 7, 0 };rank[0~n-1]为有效值，rank[n]必定为0无效值 sa[] = { 8, 3, 4, 5, 0, 6, 1, 7, 2 };sa[1~n]为有效值，sa[0]必定为n是无效值
height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 };height[2~n]为有效值
/
const int MAXN=100010;
int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
//待排序的字符串放在s数组中，从s$0$到s$n-1$,长度为n,且最大值小于m,
//除s$n-1$外的所有s$i$都大于0，r$n-1$=0
//函数结束以后结果放在sa数组中
bool cmp(int
r,int a,int b,int l){
return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int str[],int sa[],int rank[],int height[],int n,int m){
n++;
int i, j, p, *x = t1, *y = t2;
//第一轮基数排序，如果s的最大值很大，可改为快速排序
for(i = 0;i < m;i++)c[i] = 0;
for(i = 0;i < n;i++)c[x[i] = str[i]]++;
for(i = 1;i < m;i++)c[i] += c[i-1];
for(i = n-1;i >= 0;i–)sa[–c[x[i]]] = i;
for(j = 1;j <= n; j <<= 1){
p = 0;
//直接利用sa数组排序第二关键字
for(i = n-j; i < n; i++)y[p++] = i;//后面的j个数第二关键字为空的最小
for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;
//这样数组y保存的就是按照第二关键字排序的结果
//基数排序第一关键字
for(i = 0; i < m; i++)c[i] = 0;
for(i = 0; i < n; i++)c[x[y[i]]]++;
for(i = 1; i < m;i++)c[i] += c[i-1];
for(i = n-1; i >= 0;i–)sa[–c[x[y[i]]]] = y[i];
//根据sa和x数组计算新的x数组
swap(x,y);
p = 1; x[sa[0]] = 0;
for(i = 1;i < n;i++)
x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
if(p >= n)break;
m = p;//下次基数排序的最大值
}
int k = 0;
n–;
for(i = 0;i <= n;i++)rank[sa[i]] = i;
for(i = 0;i < n;i++){
if(k)k–;
j = sa[rank[i]-1];
while(str[i+k] == str[j+k])k++;
height[rank[i]] = k;
}
}
int rank[MAXN],height[MAXN];
int RMQ[MAXN];
int mm[MAXN];
int best[20][MAXN];
void initRMQ(int n){
mm[0]=-1;
for(int i=1;i<=n;i++)
mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
for(int i=1;i<=n;i++)best[0][i]=i;
for(int i=1;i<=mm[n];i++)
for(int j=1;j+(1<<i)-1<=n;j++){
int a=best[i-1][j];
int b=best[i-1][j+(1<<(i-1))];
if(RMQ[a]<RMQ[b])best[i][j]=a;
else best[i][j]=b;
}
}
int askRMQ(int a,int b){
int t;
t=mm[b-a+1];
b-=(1<<t)-1;
a=best[t][a];b=best[t][b];
return RMQ[a]<RMQ[b]?a:b;
}
int lcp(int a,int b){
a=rank[a];b=rank[b];
if(a>b)swap(a,b);
return height[askRMQ(a+1,b)];
}
char str[MAXN];
int r[MAXN];
int sa[MAXN];
int dp[20][MAXN];
void init(int n){
for(int i = 1;i <= n;i++){
dp[0][i] = sa[i];
}
for(int j = 1;j <= mm[n];j++)
for(int i = 1;i + (1<<j) - 1 <= n;i++)
dp[j][i] = min(dp[j-1][i],dp[j-1][i+(1<<(j-1))]);
}
int rmq(int x,int y){
int k = mm[y-x+1];
return min(dp[k][x],dp[k][y-(1<<k)+1]);
}

long long b[MAXN];
int main()
{
//freopen(“in.txt”,”r”,stdin);
//freopen(“out.txt”,”w”,stdout);
while(scanf(“%s”,str) == 1){
int n = strlen(str);
for(int i = 0;i < n;i++)r[i] = str[i];
r[n] = 0;
da(r,sa,rank,height,n,128);
for(int i = 1;i <= n;i++)RMQ[i] = height[i];
initRMQ(n);
init(n);
b[0] = 0;
for(int i = 1;i <= n;i++)
b[i] = b[i-1] + n - sa[i] - height[i];
int m;
scanf(“%d”,&m);
long long k;
int lastl = 0, lastr = 0;
while(m–){
scanf(“%I64d”,&k);//提交记得修改========
k = (k^lastl^lastr) + 1;
if(k > b[n]){
printf(“0 0\n”);
lastl = 0;
lastr = 0;
continue;
}
int id = lower_bound(b+1,b+n+1,k) - b;
k -= b[id-1];
int len = height[id] + k;
int ll = id;
int rr = id;
int l = id, r = n;
while(l <= r){
int mid = (l+r)/2;
if(sa[id] == sa[mid] || lcp(sa[id],sa[mid]) >= len){
rr = mid;
l = mid+1;
}
else r = mid-1;
}
int ansl = rmq(ll,rr);
int ansr = ansl + len - 1;
ansl++; ansr++;
printf(“%d %d\n”,ansl,ansr);
lastl = ansl;
lastr = ansr;
}
}
return 0;
}

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