poj3461—Oulipo(查找子串在目标串中出现次数)
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题目链接:传送门
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
3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN
Sample Output
130
直接上代码吧
#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>using namespace std;const int N = 1005000;const int M = 10100;//poj3461//计算子串在目标串中出现的次数int nt[N],pLen;char s[N],p[M];void GetNext(char* p){ int pLen = strlen(p); nt[0] = -1; int k = -1,j = 0; while (j < pLen - 1){ //p[k]表示前缀,p[j]表示后缀 if (k == -1 || p[j] == p[k]){ ++k; ++j; if (p[j] != p[k]) nt[j] = k; //之前只有这一行 else //因为不能出现p[j] = p[ nt[j ]] //所以当出现时需要继续递归,k = nt[k] = nt[nt[k]] nt[j] = nt[k]; }else{ k = nt[k]; } }}int KMP(char* s, char* p){ int i = 0,j = 0,ans = 0; int sLen = strlen(s); while (i < sLen){//找到一处匹配 if(j == pLen){ ans++; j = nt[pLen]; } //如果j = -1,或者当前字符匹配成功(即S[i] == P[j]),都令i++,j++ if (j == -1 || s[i] == p[j]){ i++; j++; }else{ //如果j != -1,且当前字符匹配失败(即S[i] != P[j]),则令 i 不变,j = nt[j] //nt[j]即为j所对应的nt值 j = nt[j]; } } if(j == pLen){ ans++; } return ans;}int main(){ int T; scanf("%d",&T); while(T--){ scanf(" %s",p); scanf( "%s",s); pLen = strlen(p); p[pLen] = '#'; GetNext(p); printf("%d\n",KMP(s,p)); } return 0;}
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