KMP算法专题-1001
来源:互联网 发布:等高线地图软件 编辑:程序博客网 时间:2024/06/11 11:36
A - Oulipo
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.
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.
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
3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN
Sample Output
130
代码:
#include<stdio.h>#include<string.h>int next[10005];char str[1000005],str2[1000005];int cnt;void getnext(int len2){int i=0,j=-1;next[0]=-1;while(i<len2){ if(j==-1||str2[i]==str2[j]) { i++;j++; if(str2[i]!=str2[j]) next[i]=j; else next[i]=next[j]; } else j=next[j];}}void kmp(int len1,int len2){int i=0,j=0;getnext(len2);while(i<len1){if(j==-1||str[i]==str2[j]) { ++i;++j; } else j=next[j]; if(j==len2) { cnt++; j=next[j]; }}}int main(){ int n; int len1,len2; scanf("%d",&n); getchar(); while(n--) { gets(str2); gets(str); len1=strlen(str); len2=strlen(str2); cnt=0; kmp(len1,len2); printf("%d\n",cnt); }}
0 0
- KMP算法专题-1001
- KMP算法专题-1002
- 【字符匹配专题】--1001 kmp
- KMP专题
- 【专题】KMP
- 【KMP专题】
- KMP专题
- 【专题】KMP
- KMP专题【完结】
- KMP专题【完结】
- KMP专题【完结】
- KMP专题小结
- KMP专题总结
- 数据结构与算法专题之串——字符串及KMP算法
- 总结_六天专题:栈和队列,树,并查集,KMP匹配算法
- KMP · 扩展KMP · Manacher 专题
- KMP · 扩展KMP · Manacher 专题
- KMP · 扩展KMP · Manacher 专题
- C++中的四种cast
- RHEL二十(管理SELINUX的安全性)
- Nginx配置文件详细说明
- 从零开始搭建SpringMVC框架以及最简单的 Hello World 实例
- android 收集已发布版本的错误信息(UncaughtExceptionHandler)
- KMP算法专题-1001
- Snail—UI学习之表视图TableView单行添加、删除和移动
- ios实现简单的计算器
- hdoj 2277 Change the ball【规律题】
- Is It A Tree?(POJ_1308)
- 数字图像学习0
- Android Native C Log
- 什么是存储过程?
- 黑马程序员—JAVA基础—io流