HDU 1686-Oulipo
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Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16570 Accepted Submission(s): 6636
题目链接:点击打开链接
Problem 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
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
题意:
题目太长了,看一眼测试测试数据,瞬间懂了,提交就对了,说明猜的没错并且题目没有坑。题目就是给你两个串然你在第二个串里找第一个串最大的匹配值,即能找到多少个和第一个串一样的串
分析:
KMP算法。
#include<stdio.h>#include<string>#include<string.h>const int INF1=10005;const int INF2=1000005;char a[INF1],b[INF2];int next[INF1];int m,n;void get_next(){ next[0] = -1; int k = -1; for (int i = 1; i < m; i++) { while (k > -1 && a[k + 1] != a[i]) { k = next[k]; } if (a[k + 1] == a[i]) { k = k + 1; } next[i] = k; }}int KMP(){ get_next(); int k = -1; int sum=0; for (int i = 0; i < n;i++ ) { while (k >-1&& a[k + 1] != b[i]) { k = next[k]; } if (a[k + 1] == b[i]) k = k + 1; if (k == m-1) { sum++; } } return sum;}int main(){ int t; scanf("%d",&t); while(t--) { scanf("%s",a); scanf("%s",b); m=strlen(a); n=strlen(b); int sum=KMP(); printf("%d\n",sum); } return 0;}
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