HDU 5769 Substring （后缀数组）

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<=301<=|S|<=10^5The 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

2aabcbbbb

Sample Output

Case #1: 3 Case #2: 3

题意

给定一个字符串，求包含字母 ​​x​​ 的不相同子串数目。

思路

这道题和之前我们所做的求字符串不同子串数目类似，唯一的不同便是限制了必须要包含字母 ​​x​​ 。

如果没有该限制，每一个后缀对结果的贡献为 len−SA[i]+1−Height[i]

我们记录第 i 位以后第一个出现字母 ​​x​​ 的位置为 last[i] ，保证不相同子串的情况下如果 last[SA[i]]+1<=len−SA[i]+1−Height[i] ，则说明当前后缀所有贡献的子串都满足条件，否则只有 len−last[SA[i]]

AC 代码

#include#define rank rankkusing 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>T; for(int ti=1; ti<=T; ti++) { char c; cin>>c; cin>>str; cout<<"Case #"<

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