2016暑假多校联合---Substring（后缀数组）

2016暑假多校联合---Substring

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

题意：输入字符x和一个字符串，求包含字符x的不同子串的个数；

思路： 后缀数组sum=length-(sa[i]+height[i])[i从1~length]  sum即为子串个数，稍作修改，用nxt[i]表示在i右侧距离i最近的字符x的坐标，则

sum=length-max(nxt[sa[i]],(sa[i]+height[i]))  [i从1~length]就是所求结果;

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e5+5;
char s[maxn];
int wa[maxn],wb[maxn],wv[maxn],wss[maxn];
int sa[maxn],ran[maxn],height[maxn];

int cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b]&&r[a+l]==r[b+l];
}

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++) wss[i]=0;
    for(i=0; i<n; i++) wss[x[i]=(int)r[i]]++;
    for(i=1; i<m; i++) wss[i]+=wss[i-1];
    for(i=n-1; i>=0; i--) sa[--wss[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++) wss[i]=0;
        for(i=0; i<n; i++) wss[wv[i]]++;
        for(i=1; i<m; i++) wss[i]+=wss[i-1];
        for(i=n-1; i>=0; i--) sa[--wss[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;
}

void callheight(char *r,int *sa,int n)
{
    int i,j,k=0;
    for(i=1;i<=n;i++)
        ran[sa[i]]=i;
    for(i=0;i<n;height[ran[i++]]=k)
    for(k?k--:0,j=sa[ran[i]-1];r[i+k]==r[j+k];k++);
    return ;
}

int main()
{
    int T;
    int Case=1;
    cin>>T;
    char x;
    while(T--)
    {
        scanf(" %c",&x);
        scanf("%s",s);
        int len=strlen(s);
        da(s,sa,len+1,130);
        callheight(s,sa,len);
        int nxt[100005];
        int tmp=len;
        long long sum=0;
        for(int i=len-1;i>=0;i--)
        {
            if(s[i]==x) tmp=i;
            nxt[i]=tmp;
        }
        for(int i=1;i<=len;i++)
        {
            sum+=(long long)(len-max(sa[i]+height[i],nxt[sa[i]]));
        }
        printf("Case #%d: %lld
",Case++,sum);
    }
    return 0;
}