# HDU 5769 Substring （后缀数组）

## 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


## AC 代码

#include<bits/stdc++.h>
#define rank rankk
using namespace std;
typedef __int64 LL;
const int MAXN = 7e5+10;

char ch[MAXN], All[MAXN];
int SA[MAXN], rank[MAXN], Height[MAXN], tax[MAXN], tp[MAXN], a[MAXN], n, m;
char str[MAXN];
int last[MAXN];
//rank[i] 第i个后缀的排名; SA[i] 排名为i的后缀位置; Height[i] 排名为i的后缀与排名为(i-1)的后缀的LCP
//tax[i] 计数排序辅助数组; tp[i] rank的辅助数组(计数排序中的第二关键字),与SA意义一样。
//a为原串
void RSort()
{
//rank第一关键字,tp第二关键字。
for (int i = 0; i <= m; i ++) tax[i] = 0;
for (int i = 1; i <= n; i ++) tax[rank[tp[i]]] ++;
for (int i = 1; i <= m; i ++) tax[i] += tax[i-1];
for (int i = n; i >= 1; i --) SA[tax[rank[tp[i]]] --] = tp[i]; //确保满足第一关键字的同时，再满足第二关键字的要求
} //计数排序,把新的二元组排序。

int cmp(int *f, int x, int y, int w)
{
return f[x] == f[y] && f[x + w] == f[y + w];
}
//通过二元组两个下标的比较，确定两个子串是否相同

void Suffix()
{
//SA
for (int i = 1; i <= n; i ++) rank[i] = a[i], tp[i] = i;
m = 127,RSort();  //一开始是以单个字符为单位，所以(m = 127)

for (int w = 1, p = 1, i; p < n; w += w, m = p)   //把子串长度翻倍,更新rank
{

//w 当前一个子串的长度; m 当前离散后的排名种类数
//当前的tp(第二关键字)可直接由上一次的SA的得到
for (p = 0, i = n - w + 1; i <= n; i ++) tp[++ p] = i; //长度越界,第二关键字为0
for (i = 1; i <= n; i ++) if (SA[i] > w) tp[++ p] = SA[i] - w;

//更新SA值,并用tp暂时存下上一轮的rank(用于cmp比较)
RSort(), swap(rank, tp), rank[SA[1]] = p = 1;

//用已经完成的SA来更新与它互逆的rank,并离散rank
for (i = 2; i <= n; i ++) rank[SA[i]] = cmp(tp, SA[i], SA[i - 1], w) ? p : ++ p;
}
//离散：把相等的字符串的rank设为相同。
//LCP
int j, k = 0;
for(int i = 1; i <= n; Height[rank[i ++]] = k)
for( k = k ? k - 1 : k, j = SA[rank[i] - 1]; a[i + k] == a[j + k]; ++ k);
//这个知道原理后就比较好理解程序
}

LL solve(char c)
{
n = strlen(str);
int la = -1;
for(int i=n-1; i>=0; i--)   //记录当前位置以后第一次出现 c 的坐标
{
if(str[i]==c)
la = i;
last[i+1]=la;
}
for(int i=0; i<n; i++)
a[i+1]=str[i];
Suffix();
LL ans=0;
for(int i=1; i<=n; i++)
if(last[SA[i]]!=-1)
ans+= n+1 - max(SA[i]+Height[i],last[SA[i]]+1);
return ans;
}

int main()
{
ios::sync_with_stdio(false);
int T;
cin>>T;
for(int ti=1; ti<=T; ti++)
{
char c;
cin>>c;
cin>>str;
cout<<"Case #"<<ti<<": "<<solve(c)<<endl;
}
return 0;
}