HDU:1686 Oulipo(KMP包含字符串(不是分割))
2016-08-16 21:39
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Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11360 Accepted Submission(s): 4458
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
[align=left]Sample Output[/align]
1 3 0
[align=left]Source[/align]
华东区大学生程序设计邀请赛_热身赛
[align=left]Recommend[/align]
lcy | We have carefully selected several similar problems for you: 1358 3336 3746 2203 3068
解题思路:给你一个字符串s2,一个字符串s1,问s1中包含了多少个s2(注意是包含不是分割,如上面的例子AZAZAZA 和AZA,如果是分割的话答案是2,如果是包含的话答案是3,包含的话注意匹配成功后k怎么处理,看代码)
代码如下:
#include <cstdio> #include <cstring> char s1[1000010]; char s2[10010]; int next[10010]; int ans,size1,size2; void makenext() { int k=0; memset(next,0,sizeof(next)); for(int i=1;i<size2;i++) { while(k>0&&s2[k]!=s2[i]) { k=next[k-1]; } if(s2[k]==s2[i]) { k++; } next[i]=k; } } void kmp() { int k=0; for(int i=0;i<size1;i++) { while(k>0&&s2[k]!=s1[i]) { k=next[k-1]; } if(s2[k]==s1[i]) { k++; } if(k==size2) { ans++; k=next[k-1];//注意这里,最后一个点当做不相等吧,然后就得滑动s2串了,符合题意的包含,不是分割 } } } int main() { int t; scanf("%d",&t); while(t--) { scanf("%s",s2); scanf("%s",s1); size1=strlen(s1); size2=strlen(s2); makenext(); ans=0; kmp(); printf("%d\n",ans); } return 0; }
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