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
- Thief in a Shop
题意: 问n个物品选出K个可以拼成的体积有哪些. 解法: 多项式裸题,注意到本题中 $A(x)^K$ 的系数会非常大,采用NTT优于FFT. NTT 采用两个 $2^t+1$ 质数,求原根 $g_n$ ...
- 总结open与fopen的区别
https://www.zybuluo.com/yiltoncent/note/87461 参考链接1 参考链接2 对于这两个名字很类似的函数,对于很多初学者来说,不容易搞清楚它们有什么不同,只知道按 ...
- 解决 'chromedriver' executable needs to be in PATH.'报错
试了把chromedriver.exe放到chrome安装文件下,python安装文件下,然后把路径配到path里,均无用. 最后是修改函数调用得以解决: from selenium import w ...
- C++泛型编程之函数模板
泛型语义 泛型(Generic Programming),即是指具有在多种数据类型上皆可操作的含意.泛型编程的代表作品 STL 是一种高效.泛型.可交互操作的软件组件. 泛型编程最初诞生于 C++中, ...
- 7 二分搜索树的原理与Java源码实现
1 折半查找法 了解二叉查找树之前,先来看看折半查找法,也叫二分查找法 在一个有序的整数数组中(假如是从小到大排序的),如果查找某个元素,返回元素的索引. 如下: int[] arr = new in ...
- uoj#282. 长度测量鸡(构造)
传送门 打表题--只有\(n\leq 3\)有解否则无解→_→ 或者严格证明的话是这样,因为算上端点一共\(n+1\)个点,共\(\frac{n(n+1)}{2}\)个点对,所以点对之间两两距离不相等 ...
- route(2018.10.24)
建出最短路图之后\(topsort\)即可. 具体思路: 先用\(dijkstra\)算法在原图中跑出\(1\)号点到\(i\)号节点的最短距离\(dist_1(i)\),将所有边反向后用\(dijk ...
- 黑马MySQL数据库学习day02 表数据CRUD 约束CRUD
/* 基础查询练习: 1.字段列表查询 当查询全部字段时,一种简便方式,使用*代替全部字段(企业中不推荐使用) 2.去除重复行 DISTINCT,注意修饰的是行,也就是整个字段列表,而不是单个字段. ...
- MySQL习题1 一对多实例 产品和分类
/* 需求:建立产品和分类表 1.查询每种分类的产品数量,没有产品的分类也要统计.(cname,quantity) 2.根据分类名称查询分类中的所有产品 */ -- ----------------- ...
- SLAM学习资料整理(转)
原文出处:http://www.cnblogs.com/wenhust/p/5942893.html 书籍: 1.必读经典 Thrun S, Burgard W, Fox D. <Probabi ...