hdu4622(后缀数组+ST算法)

Reincarnation

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1305    Accepted Submission(s): 448

Problem Description

Now you are back,and have a task to do:
Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.
And you have some query,each time you should calculate f(s[l...r]), s[l...r] means the sub-string of s start from l end at r.

 

Input

The first line contains integer T(1<=T<=5), denote the number of the test cases.
For each test cases,the first line contains a string s(1 <= length of s <= 2000).
Denote the length of s by n.
The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.

 

Output

For each test cases,for each query,print the answer in one line.

 

Sample Input

2bbaba53 42 22 52 41 4baaba53 33 41 43 55 5

 

Sample Output

3175813851
Hint
      I won't do anything against hash because I am nice.Of course this problem has a solution that don't rely on hash.
 

 

Source

2013 Multi-University Training Contest 3

 

Recommend

zhuyuanchen520

         本题要求给定字符区间内不同串的个数。由于看了那篇大牛的论文，知道是后缀数组的题目。

       首先求后缀数组sa[],height[].由于本题所查询区间是原字符串的一部分，这给题目带来难度。我们首先看最简单的模型。每增加一个后缀最多可增加len-sa[i]+1个前缀，如果i从1到len进行的话，最终即可得原串中所有的字串，但是应该去重。每添加一个后缀i,其对应的前缀最多增加len-sa[i]+1，这其中有些是已经存在的height[i]个(其与排名在他前面一位的最长公共前缀)。而本题要做的是给定区间的查询问题。想了几乎一天，也没能想到好点的方法，于是决定枚举排名，假设查询区间是[left,right]，每遇到一个下标位于该区间的，就看与他同区间的且在他前面的与它的公共前缀的长度（任意两个后缀的最长公共前缀问题可以根据所求的height数组利用线段树进行预处理），然后注意有没有超过该区间即可。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;//*****************************************************************const int MAXN= 2000+100;const int Mpow=12;//保证2^(Mpow)>MAXN即可char str[MAXN];//待处理字符串int sa[MAXN];//求得的后缀数组int wa[MAXN],wb[MAXN],wv[MAXN],wh[MAXN];int cmp(int *r,int a,int b,int l){return r[a]==r[b]&&r[a+l]==r[b+l];}//求后缀数组sa[],下标1到n-1(此处n=strlen(str)+1)有效后缀//将str的n个后缀从小到大进行排序之后把排好序的后缀的开头位置顺次放入sa中。//保证Suffix(sa[i])<Suffix(sa[i+1])//1<=i<n,sa[0]存放人为添加在末尾的那个最小的后缀//倍增算法的时间复杂度为O(nlogn)//倍增算法的空间复杂度都是O(n)void da(char *r,int *sa,int n,int m){int i,j,p,*x=wa,*y=wb,*t;for(i=0;i<m;i++) wh[i]=0;for(i=0;i<n;i++) wh[x[i]=r[i]]++;for(i=1;i<m;i++) wh[i]+=wh[i-1];for(i=n-1;i>=0;i--) sa[--wh[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++) wh[i]=0;for(i=0;i<n;i++) wh[wv[i]]++;for(i=1;i<m;i++) wh[i]+=wh[i-1];for(i=n-1;i>=0;i--) sa[--wh[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;}int Rank[MAXN],height[MAXN];//定义height[i]=suffix(sa[i-1])和suffix(sa[i])的最长公//共前缀，也就是排名相邻的两个后缀的最长公共前缀//任意两个起始位置为i,j(假设Rank[i]<Rank[j])的后缀的最长公共前缀//为height[Rank[i]+1]、height[Rank[i]+2]…height[Rank[j]]的最小值void calheight(char *r,int *sa,int n){int i,j,k=0;for(i=1;i<=n;i++) Rank[sa[i]]=i;for(i=0;i<n;height[Rank[i++]]=k)for(k?k--:0,j=sa[Rank[i]-1];r[i+k]==r[j+k];k++);return;}//*****************************************************************int Mindp[MAXN][Mpow];inline int Min(int a,int b){return a<b?a:b;}inline void init(int n){for(int i=1;i<=n;i++){Mindp[i][0]=height[i];}}//预处理过程，时间复杂度为O(N*log(N))//ST算法求区间最值//dp[i][j]表示区间[i,i+2^j-1]最值//求dp[i][j]时将其分成dp[i][j-1],dp[i+2^(j-1)][j-1]//[i,i+2^j-1]=[i,i+2^(j-1)-1]+[i+2^(j-1),i+2^(j-1)+2^(j-1)-1]inline void Rmp_ST(int n){int l,s;init(n);for(l=1;l<=16;l++){for(s=1;s<=n;s++){if(s+(1<<l)-1<=n){Mindp[s][l]=Min(Mindp[s][l-1],Mindp[s+(1<<(l-1))][l-1]);}}}}inline int Rmp_ST_query(int s,int e){int Min_ans,Max_ans;int k=(int)(log(1.0*e-s+1)/log(2.0));return Min(Mindp[s][k],Mindp[e-(1<<(k))+1][k]);}int len;//整个字符串的长度int Query(int left,int right){int i,ans=0,pre=-1,tot;int op=0;int most=right-left+1;//int tmp;for(i=1;i<=len&&op<most;i++){if(sa[i]>=left&&sa[i]<=right){//tmp=right-sa[i]+1;if(-1==pre){pre=sa[i];//前一个的下标//printf("pre=%d i=%d  ",pre,i);ans+=(right-pre+1);//printf("ans=%d \n",ans);}else {//tot=Min(right-sa[i],right-sa[pre])+1;tot=right-sa[i]+1;//最多包含这么多个前缀int add,tt=Rmp_ST_query(Rank[pre]+1,i);if(pre+tt<=right&&sa[i]+tt<=right){//printf("pre+tt<=right&&sa[i]+tt<=right\n");add=(tot-tt);}else if(pre+tt<=right){//printf("pre+tt<=right&&sa[i]+tt>right\n");add=0;}else if(sa[i]+tt<=right){//printf("pre+tt>right&&sa[i]+tt<=right\n");add=tot-(right-pre+1);}ans+=add;//printf("pre=%d  sa[%d]=%d  tt=%d  add=%d  tot=%d  ans=%d\n",pre,i,sa[i],tt,add,tot,ans);pre=sa[i];}op++;}}return ans;}int main(){int cas,n,q,left,right;//freopen("in.txt","r",stdin);cin>>cas;while(cas--){scanf("%s",str);scanf("%d",&q);len=strlen(str);da(str,sa,len+1,200);calheight(str,sa,len);Rmp_ST(len);while(q--){scanf("%d%d",&left,&right);left--;right--;//printf("left=%d,right=%d\n",left,right);//printf("************%d\n\n",Query(left,right));printf("%d\n",Query(left,right));}}return 0;}