POJ 3461 Oulipo （kmp模板題）
題目連結：POJ 3461
| Describe: |
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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:
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' 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 |
| Input: |
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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:
|
| 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 |
題目大意：
若干樣例，每個樣例第一行給出一個單詞，第二行給出一個字串，要求輸出，字串中有多少個給定的單詞。
解題思路：
比較裸的KMP，不過有一點需要根據kmp得到的next陣列進行優化，在我們匹配完一次單詞以後，如果所給單詞從頭匹配就會TLE，所以此時讓下標j=next[j-1]，道理很簡單，同樣是根據字首字尾，公共的部分已經匹配過，不需要重複匹配。
AC程式碼：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 char str[1000010]; // 輸入的字串 6 char pattern[10010]; // 輸入的單詞，相當於模式串 7 int next[10010]; 8 int n1,n2; // 兩個串的長度 9 // 常規next陣列 10 void getnext() 11 { 12 memset(next,0,sizeof(next)); 13 int i,j; 14 for(i=1,j=0; i < n2; ) 15 { 16 if(pattern[i] == pattern[j]) { 17 next[i] = next[i-1]+1; 18 i++;j++; 19 } else { 20 if(j != 0) 21 { 22 j = next[j-1]; 23 } else next[i++] = 0; 24 } 25 } 26 } 27 // 常規kmp 28 int kmp() 29 { 30 int ans = 0; // 記錄出現次數 31 int i,j; 32 for(i=0,j=0; i <= n1; ) // 迴圈條件注意，我這種寫法要有等號 33 { 34 if(j == n2) 35 { 36 j = next[j-1]; // 重點，根據next陣列優化 37 ans++; 38 if(i == n1) break; 39 } 40 if(str[i] == pattern[j]) 41 { 42 i++; 43 j++; 44 } else { 45 if(j != 0) j = next[j-1]; 46 else i++; 47 } 48 } 49 return ans; 50 } 51 int main() 52 { 53 int T; 54 scanf("%d",&T); 55 while(T--) 56 { 57 scanf("%s%s",pattern,str); 58 n1 = strlen(str); 59 n2 = strlen(pattern); 60 getnext(); 61 printf("%d\n",kmp()); 62 } 63 return 0; 64 }