POJ 3461 kmp
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 40168 | Accepted: 16135 |
Description
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
Source
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char a[];
char b[];
int p[];
int la,lb;
int j;
int ans;
inline void makep()
{ j=;
for(int i=;i<la;i++)
{
while(j>&&a[i]!=a[j])
j=p[j];
if(a[i]==a[j])
j++;
p[i]=j;
}
return ;
}
inline void KMP()
{
j=;
ans=;
for(int i=;i<lb;i++)
{
while(b[i]!=a[j]&&j>)
j=p[j-];
if(b[i]==a[j])
j++;
if(j==la)
{
ans++;
j=p[j-];
} }
printf("%d\n",ans);
return ;
} int main()
{
int n;
scanf("%d",&n); for(int i=;i<=n;i++)
{
memset(p,,sizeof(p));
scanf("%s%s",a,b);
la=strlen(a);lb=strlen(b);
makep();
KMP();
} return ;
}
TLE
AC
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
using namespace std; int n,la,lb;
int next[];
char a[],b[]; void getnext()
{
int j=-;
next[]=-;
for(int i=;i<lb;i++)
{
if(j!=- && b[j+]!=b[i]) j=next[j];
if(b[j+]==b[i]) j++;
next[i]=j;
}
} int kmp()
{
int ans=;
int j=-;
for(int i=;i<la;i++)
{
if(j!=- && a[i]!=b[j+]) j=next[j];
if(b[j+]==a[i]) j++;
if(j==lb-)
{
ans++;
j=next[j];
}
}
return ans;
} int main()
{
scanf("%d",&n);
for(int u=;u<=n;u++)
{
scanf("%s %s",b,a);
la=strlen(a);
lb=strlen(b);
getnext();
printf("%d\n",kmp());
}
}
AC
POJ 3461 kmp的更多相关文章
- POJ 3461 kmp 应用
题意:求匹配串在文本中出现次数,KMP应用,理解了就OK了,每次匹配成功就累加次数,开始的时候超时, 由于在处理每次成功的时候让i=i-len2+1,相当于回溯了,后来一想,本次成功,相当于" ...
- HDU 1686 Oulipo / POJ 3461 Oulipo / SCU 2652 Oulipo (字符串匹配,KMP)
HDU 1686 Oulipo / POJ 3461 Oulipo / SCU 2652 Oulipo (字符串匹配,KMP) Description The French author George ...
- POJ 3461 Oulipo[附KMP算法详细流程讲解]
E - Oulipo Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- POJ 3461 Oulipo
E - Oulipo Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- POJ 3461 Oulipo(乌力波)
POJ 3461 Oulipo(乌力波) Time Limit: 1000MS Memory Limit: 65536K [Description] [题目描述] The French autho ...
- (KMP)Oulipo -- poj --3461
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=92486#problem/B http://poj.org/problem?id=3461 ...
- POJ 3461 Oulipo 【KMP统计子串数】
传送门:http://poj.org/problem?id=3461 Oulipo Time Limit: 1000MS Memory Limit: 65536K Total Submission ...
- POJ - 3461 (kmp)
题目链接:http://poj.org/problem?id=3461 Oulipo Time Limit: 1000MS Memory Limit: 65536K Total Submissio ...
- POJ 3080 Blue Jeans、POJ 3461 Oulipo——KMP应用
题目:POJ3080 http://poj.org/problem?id=3080 题意:对于输入的文本串,输出最长的公共子串,如果长度相同,输出字典序最小的. 这题数据量很小,用暴力也是16ms,用 ...
随机推荐
- Java笔记(四)
13. 集合框架: 集合中存储的都是对象的引用(地址) 迭代器:集合的取出元素的方式 import java.util.ArrayList; import java.util.Iterator; pu ...
- MongoDB 2.6复制集单节点部署(三)
MongoDB在单节点中也可以做复制集,但是仅限于测试实验,最大的好处就是部署方便快速,可以随便添加新节点,节省资源.在这里我使用的是MongoDB 2.6版本进行复制集实验(但MongoDB配置文件 ...
- tyvj1940创世纪——贪心(基环树)
题目:http://www.joyoi.cn/problem/tyvj-1940 基环树的样子,看了书上的讲解,准备写树上DP,然后挂了: #include<iostream> #incl ...
- SPOJ(后缀数组求不同子串个数)
DISUBSTR - Distinct Substrings Given a string, we need to find the total number of its distinct subs ...
- HDU2087(KMP入门题)
剪花布条 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submis ...
- htop 详细功能使用简介
一.htop 简介 This is htop, an interactive process viewer for Linux. It is a text-mode application (for ...
- win10文件夹或文件已在另一程序中打开
我们在对文件或文件夹进行删除.移动.重命名等操作时,系统可能提示“操作无法完成,因为其中的文件夹已在另一程序中打开,请关闭该文件或文件夹,然后重试.”,遇到这种情况我们应该怎么办呢?请看下文. 方法/ ...
- git搭建私有仓库
git gui参考 https://ask.helplib.com/git/post_1004941
- sql之视图、触发器、函数、存储过程、事务
视图 # 视图也是一张表,但在data文件里只有表结构,没有表数据 # 不建议使用,扩展性差,程序需改变时,依赖的视图也要改变 # 视图牵涉到多张表时,视图中的记录不能修改. create view ...
- 基于Jenkins自动构建系统开发
1 绪论 1.1 课题的研究背景 随着IT行业的不断发展,软件开发的复杂度也随着不断提高.与此同时,软件的开发团队也越来越庞大,而如何更好地协同整个团队进行高效准确的工作,从而确保软件开发的质量已经 ...