HDU 5769-Substring(后缀数组-不相同的子串的个数)

Substring

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 973    Accepted Submission(s): 393

Problem Description

?? is practicing his program skill, and now he is given a string, he has to calculate the total number of its distinct substrings. 
But ?? thinks that is too easy, he wants to make this problem more interesting. 
?? likes a character X very much, so he wants to know the number of distinct substrings which contains at least one X. 
However, ?? is unable to solve it, please help him.

 

Input

The first line of the input gives the number of test cases T;T test cases follow. 
Each test case is consist of 2 lines: 
First line is a character X, and second line is a string S. 
X is a lowercase letter, and S contains lowercase letters(‘a’-‘z’) only.

T<=30 
1<=|S|<=10^5 
The sum of |S| in all the test cases is no more than 700,000.

 

Output

For each test case, output one line containing “Case #x: y”(without quotes), where x is the test case number(starting from 1) and y is the answer you get for that case.

 

Sample Input

2 a abc b bbb

 

Sample Output

Case #1: 3 Case #2: 3
Hint
In first case, all distinct substrings containing at least one a: a, ab, abc. In second case, all distinct substrings containing at least one b: b, bb, bbb.
 

 

Author

FZU

 

Source

2016 Multi-University Training Contest 4

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5808 5807 5806 5805 5804 

题目意思：

有一个子串X，一个母串str，求str中不相同的子串个数且至少含有一个X串。

解题思路：

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#define maxn 1000010using namespace std;//以下为倍增算法求后缀数组int wa[maxn],wb[maxn],wv[maxn],Ws[maxn],nxt[maxn];int sa[maxn],Rank[maxn],height[maxn];char str[maxn],X[maxn];int cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(const char *r,int *sa,int n,int m){    int i,j,p,*x=wa,*y=wb,*t;    for(i=0; i<m; i++) Ws[i]=0;    for(i=0; i<n; i++) Ws[x[i]=r[i]]++;    for(i=1; i<m; i++) Ws[i]+=Ws[i-1];    for(i=n-1; i>=0; i--) sa[--Ws[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++) Ws[i]=0;        for(i=0; i<n; i++) Ws[wv[i]]++;        for(i=1; i<m; i++) Ws[i]+=Ws[i-1];        for(i=n-1; i>=0; i--) sa[--Ws[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++;    }}//求height数组void calheight(const 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 slove(int n){    int sum=0;    for(int i=1; i<=n; i++)        sum+=n-sa[i]-height[i];    return sum;}int main(){    int t,ca=0;    scanf("%d",&t);    while(t--)    {        scanf("%s%s",X,str);        int n=strlen(str);        int temp=n;        long long ans=0;        for(int i=n-1; i>=0; i--)        {            if(str[i]==X[0])            {                temp=i; cout<<temp<<"===";            }           else cout<<temp<<"***";            nxt[i]=temp;        }        da(str,sa,strlen(str)+1,130);        calheight(str,sa,strlen(str));        for(int i=1; i<=n; i++)            // printf("%d %d\n",sa[i],height[i]);            // printf("%d\n",slove(strlen(str)));            ans+=n-max(nxt[sa[i]],sa[i]+height[i]);        printf("Case #%d: %I64d\n",++ca,ans);    }    return 0;}/**2aabcbbbb**/