题目链接:http://poj.org/problem?id=3461

Oulipo
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 49608   Accepted: 19699

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

 
题目大意:输入t,代表t组样例,每个样例有两个字符串,求第一个字符串在第二个字符串中出现的次数
思路:可以说是kmp的裸题,注意的是kmp好多题目都卡cin,都要用scanf才不会超时
看代码:
#include<iostream>
#include<string.h>
#include<map>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<cmath>
#include<ctype.h>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=;
const int maxn=1e6+;
const int maxk=5e3+;
const int maxx=1e4+;
const ll maxe=+;
#define INF 0x3f3f3f3f3f3f
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
int next[maxn];
void cal_next(char a[])
{
int len=strlen(a);
next[]=-;
int k=-;
for(int i=;i<len;i++)
{
while(k>-&&a[k+]!=a[i])
{
k=next[k];
}
if(a[k+]==a[i]) k++;
next[i]=k;
}
}
void kmp(char a[],char b[])
{
int k=-,sum=;
int len1=strlen(a);
int len2=strlen(b);
for(int i=;i<len2;i++)
{
while(k>-&&a[k+]!=b[i])
{
k=next[k];
}
if(a[k+]==b[i]) k++;
if(k==len1-)
{
sum++;
k=next[k];
}
}
printf("%d\n",sum);
//cout<<sum<<endl;
}
int main()
{
//string a,b;
char a[maxn],b[maxn];
int t;
//cin>>t;
scanf("%d",&t);
while(t--)
{
//cin>>a>>b;
scanf("%s",a);
getchar();
scanf("%s",b);
cal_next(a);
kmp(a,b);
}
return ;
}

POJ - 3461 (kmp)的更多相关文章

  1. Oulipo POJ - 3461(kmp,求重叠匹配个数)

    Problem Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, w ...

  2. poj2406 Power Strings(kmp)

    poj2406 Power Strings(kmp) 给出一个字符串,问这个字符串是一个字符串重复几次.要求最大化重复次数. 若当前字符串为S,用kmp匹配'\0'+S和S即可. #include & ...

  3. POJ 2406 Power Strings(KMP)

    Description Given two strings a and b we define a*b to be their concatenation. For example, if a = & ...

  4. POJ题目(转)

    http://www.cnblogs.com/kuangbin/archive/2011/07/29/2120667.html 初期:一.基本算法:     (1)枚举. (poj1753,poj29 ...

  5. LightOJ 1258 Making Huge Palindromes(KMP)

    题意 给定一个字符串 \(S\) ,一次操作可以在这个字符串的右边增加任意一个字符.求操作之后的最短字符串,满足操作结束后的字符串是回文. \(1 \leq |S| \leq 10^6\) 思路 \( ...

  6. codeM编程大赛E题 (暴力+字符串匹配(kmp))

    题目大意:S(n,k)用k(2-16)进制表示1-n的数字所组成的字符串,例如S(16,16)=123456789ABCDEF10: 解题思路: n最大50000,k最大100000,以为暴力会超时. ...

  7. 经典串匹配算法(KMP)解析

    一.问题重述 现有字符串S1,求S1中与字符串S2完全匹配的部分,例如: S1 = "ababaababc" S2 = "ababc" 那么得到匹配的结果是5( ...

  8. URAL 1732 Ministry of Truth(KMP)

    Description In whiteblack on blackwhite is written the utterance that has been censored by the Minis ...

  9. Leetcode28--->字符串的匹配(KMP)

    题目: 题目的本质是给定两个字符串str1,str2,求str1中的str2串开始的地方,即字符串的匹配,KMP算法 思路:时间复杂度为O(m + n),空间复杂度为O(n),原串的长度为m,子串的长 ...

随机推荐

  1. Poj 1936,3302 Subsequence(LCS)

    一.Description(3302) Given a string s of length n, a subsequence of it, is defined as another string ...

  2. [快手(AAuto)学习笔记]如何让程序在运行时请求管理员权限(UAC)

    作者:ffsystem 作为(糟糕的)程序猿,习惯写代码解决一些简单事务.正常用批处理就能解决大部分工作,复杂一点用AutoIt 3. 有时候要分发给别人,就需要一个界面.外行你程序写得如何他看不懂, ...

  3. MyEclipse 手动安装Velocity 编辑器

    最近项目有使用Velocity 模板引擎,从而会用到*.VM页面!Myeclipse打开VM页面字体一片漆黑,哪有JSP那样看起来舒服(个人感觉)!为了解决这一问题就要安装Velocity编辑器,安装 ...

  4. C#的闭包

    简单的理解:闭包变量是把局部变量的作用域扩展到回调函数,发生在匿名方法注册到委托上,而匿名方法中使用外部的局部变量 说什么都不如图示那么容易明白啊 先看C#的源码 class Program { st ...

  5. SVD实例

    >> X = rand(5,7) X = 0.9797 0.1365 0.6614 0.5828 0.2259 0.2091 0.5678 0.2714 0.0118 0.2844 0.4 ...

  6. 5.JasperReports学习笔记5-其它数据生成动态的报表(WEB)

    转自:http://www.blogjava.net/vjame/archive/2013/10/12/404908.html 一.空数据(Empty Datasources) 就是说JRXML文件里 ...

  7. BluetoothFindRadioClose 函数

    BOOL BluetoothFindRadioClose( HBLUETOOTH_RADIO_FIND hFind );关闭与查找蓝牙无线电相关的枚举句柄.参数: hFind Enumeration ...

  8. struts 文件上传示例

    import java.io.File; import java.io.FileInputStream; import java.io.FileOutputStream; import java.io ...

  9. 基本算法思想之穷举法(C++语言描述)

    穷举算法(Exhaustive Attack method)是最简单的一种算法,其依赖于计算机的强大计算能力来穷尽每一种可能性,从而达到求解问题的目的.穷举算法效率不高,但是适应于一些没有规律可循的场 ...

  10. error C2039: “addTextureMesh”: 不是“pcl::visualization::PCLVisualizer”的成员

    error C2039: "addTextureMesh": 不是"pcl::visualization::PCLVisualizer"的成员 PCL 1.6 ...