• 【POJ3461】【KMP】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.

    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

    【分析】
    模板
     1 /*
     2 宋代李之仪
     3 卜算子·我住长江头
     4 我住长江头,君住长江尾。日日思君不见君,共饮长江水。 
     5 此水几时休,此恨何时已。只愿君心似我心,定不负相思意。
     6 */
     7 #include <iostream>
     8 #include <cstdio>
     9 #include <algorithm>
    10 #include <cstring>
    11 #include <vector>
    12 #include <utility>
    13 #include <iomanip>
    14 #include <string>
    15 #include <cmath>
    16 #include <queue>
    17 #include <assert.h>
    18 #include <map>
    19 #include <ctime>
    20 #include <cstdlib>
    21 #include <stack>
    22 #define LOCAL
    23 const int MAXN = 1000000 + 10;
    24 const int INF = 100000000;
    25 const int SIZE = 450;
    26 const int MAXM = 1000000 + 10;
    27 const int maxnode =  0x7fffffff + 10;
    28 using namespace std;
    29 int l1, l2;
    30 char a[MAXN], b[MAXN];
    31 int next[20000];//不用开太大了.. 
    32 void getNext(){
    33      //初始化next数组 
    34      next[1] = 0;
    35      int j = 0;
    36      for (int i = 2; i <= l1; i++){
    37          while (j > 0 && a[j + 1] != a[i]) j = next[j];
    38          if (a[j + 1] == a[i]) j++;
    39          next[i] = j;
    40      }
    41      return;
    42 }
    43 int kmp(){
    44     int j = 0, cnt = 0;
    45     for (int i = 1; i <= l2; i++){
    46         while (j > 0 && a[j + 1] != b[i]) j = next[j];
    47         if (a[j + 1] == b[i]) j++;
    48         if (j == l1){
    49            cnt++;
    50            j = next[j];//回到上一个匹配点 
    51         }
    52     }
    53     return cnt;
    54 }
    55 
    56 void init(){
    57      scanf("%s", a + 1);
    58      scanf("%s", b + 1);
    59      l1 = strlen(a + 1);
    60      l2 = strlen(b + 1);
    61 }
    62 
    63 int main(){
    64     int T;
    65     
    66     scanf("%d", &T);
    67     while (T--){
    68           init();
    69           getNext();
    70           printf("%d
    ", kmp());
    71     }
    72     /*scanf("%s", a + 1);
    73     l1 = strlen(a + 1);
    74     getNext();
    75     for (int i = 1; i <= l1; i++) printf("%d ", next[i]);*/
    76     return 0;
    77 }
    View Code
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  • 原文地址:https://www.cnblogs.com/hoskey/p/4333362.html
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