HDU 5769-Substring(后缀数组-不相同的子串的个数)，hdu5769-substring

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 HintIn 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 1000010

using 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;
}
/**
2
a
abc
b
bbb
**/