这部分主要涉及了对AC自动机的理解，和KMP的理解
注意点：模板～！！
AC自动机模板 和KMP的模板
字典树的建立
KMP next 数组也是非常重要
next数组性质：
1、根据KMP的next函数的性质，已知字符串t第K个字符的next[k],那么d=k-next[k],如果k%d==0,那么t[1……k]最多可均匀的分成k/d份。也就是可以生成一个长度为d的重复度为k/d的字串
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
- One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
- One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
简单的KMP匹配
#include
#include
#include
#include
#include
#include
#define MOD 1000000007
typedef long long ll;
using namespace std;
const int N = 1000005;
void kmp_pre(char x[],int m,int next[])
{ int i,j; j=next[0]=-1; i=0;
while(i=m)
{ ans++; j=next1[j]; }
}
return ans;
}
char w[10005];
char s[1000005];
int t;
int main(){
// freopen("data.txt","r",stdin);
scanf("%d",&t);
while(~scanf("%s %s",w,s)){
printf("%d\n",KMP_Count(w,strlen(w),s,strlen(s)));
}
return 0;
}
B - 统计难题_字典树
Ignatius最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的前缀).
Input输入数据的第一部分是一张单词表,每行一个单词,单词的长度不超过10,它们代表的是老师交给Ignatius统计的单词,一个空行代表单词表的结束.第二部分是一连串的提问,每行一个提问,每个提问都是一个字符串.
注意:本题只有一组测试数据,处理到文件结束.
Output对于每个提问,给出以该字符串为前缀的单词的数量.
Sample Input
banana
band
bee
absolute
acm
ba
b
band
abc
Sample Output
2
3
1
0
字典树入门题，先建一个字典树，然后遍历求解就好。
#include
#include
#include
#include
#include
#include
#define MOD 1000000007
typedef long long ll;
using namespace std;
const int N = 1000005;
struct Trie
{
int v;//v可以根据实际情况任意变化，在这里v是每个字母的次数;
Trie *next[26];
};
Trie root;
void createTrie(char *str)//建立字典树;
{
int len=strlen(str);
Trie *p=&root,*q;
for(int i=0;inext[id]==NULL)
{
q=(Trie *)malloc(sizeof(root));//申请一块新内存;
q->v=1;//v遇到新字母每一层都初始化为1;
for(int j=0;j<26;j++)
q->next[j]=NULL;
p->next[id]=q;
p=p->next[id];
}
else
{
p->next[id]->v++;//当第一个输入的字符串和后面又相等的时候，v++;
p=p->next[id];
}
}
}
int findTrie(char *str)//在字典树里查询;
{
int len=strlen(str);
Trie *p=&root;
for(int i=0;inext[id];
if(p==NULL)
return 0;
}
return p->v;//相同的字母个数;
}
char str[1000000];
int t;
int main(){
// freopen("data.txt","r",stdin);
while(gets(str) && str[0]!='\0')
{
createTrie(str);
}
while(cin>>str)
{
int ans=findTrie(str);
cout<
C - Keywords Search
In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
InputFirst line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.
OutputPrint how many keywords are contained in the description. Sample Input
1
5
she
he
say
shr
her
yasherhs
Sample Output
3
入门AC自动机
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
D - Power Strings
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input
abcd
aaaa
ababab
.
Sample Output
1
4
3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
这是一个kmp比较神奇的例子
根据KMP的next函数的性质，已知字符串t第K个字符的next[k],那么d=k-next[k],如果k%d==0,那么t[1……k]最多可均匀的分成k/d份。也就是可以生成一个长度为d的重复度为k/d的字串。
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.
一个简单的BFS
。。。
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
G - Period
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A
K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.
Input
The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.
Output
For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
Sample Input
3
aaa
12
aabaabaabaab
0
Sample Output
Test case #1
2 2
3 3
Test case #2
2 2
6 2
9 3
12 4
与D题原理相同
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
