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 = a 1a 2 ...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
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 = a 1a 2 ...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| ≤ 10 5) with only lowercase letters.
Next line contains a postive integer q(1 ≤ q ≤ 10 5), 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 < 2 63, “⊕” denotes exclusive or)
Each test case begins with a line containing a string s(|s| ≤ 10 5) with only lowercase letters.
Next line contains a postive integer q(1 ≤ q ≤ 10 5), 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 < 2 63, “⊕” 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
aaa 4 0 2 3 5
Sample Output
1 1 1 3 1 2 0 0
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 long
using 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;
}