BNUOJ 3580 Oulipo

Oulipo

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3461
64-bit integer IO format: %lld Java class name: Main

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

解题：KMP，看不懂，网上找了点代码，研究研究，这份代码写得很俊啊！

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 using namespace std;
 5 const int MaxN = 1000010;
 6 char word[MaxN/10], txt[MaxN];
 7 int next[MaxN/10];
 8 void KMP_next(char b[], int pre[]) {
 9     int n = strlen(b), k;
10     pre[0] = -1;
11     k = -1;
12     for(int i = 1; i < n; i++) {
13         while(k > -1 && b[k+1] != b[i]) k = pre[k];
14         if(b[k+1] == b[i]) k++;
15         pre[i] = k;
16     }
17 }
18 
19 int main() {
20     int n;
21     scanf("%d%*",&n);
22     while(n--) {
23         gets(word);
24         gets(txt);
25         KMP_next(word, next);
26         int cnt = 0, len = strlen(word);
27         for(int i = 0, j = -1; txt[i]; ++i) {
28             while(j > -1 && word[j+1] != txt[i]) j = next[j];
29             if(word[j+1] == txt[i]) j++;
30             if(j == len-1) {
31                 cnt++;
32                 j = next[j];
33             }
34         }
35         printf("%d
", cnt);
36     }
37     return 0;
38 }

View Code

Source

BAPC 2006 Qualification

貌似这样写。。。。更优化，但是对此题而言，无用

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 const int maxn = 1000005;
 6 int fail[maxn];
 7 void getNext(const char *pStr, int *nextArr) {
 8     int i = 0, j = -1, pLen = strlen(pStr);
 9     nextArr[i] = j;
10     while (i < pLen) {
11         if (pStr[++i] != pStr[++j]) {
12             nextArr[i] = j;
13             while (j != -1 && pStr[i] != pStr[j]) j = nextArr[j];
14         } else nextArr[i] = nextArr[j];
15     }
16 }
17 char word[maxn],text[maxn];
18 int main(){
19     int n,ret;
20     scanf("%d",&n);
21     while(n--){
22         scanf("%s %s",word,text);
23         getNext(word,fail);
24         for(int i = ret = 0,j = 0; text[i]; ++i){
25             while(j != -1 && word[j] != text[i]) j = fail[j];
26             if(!word[++j]) ret++;
27         }
28         printf("%d
",ret);
29     }
30     return 0;
31 }

View Code

整理以后的，可以选择开启所谓的优化

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 const int maxn = 1000005;
 6 int fail[maxn];
 7 char word[maxn],text[maxn];
 8 void getFail() {
 9     fail[0] = -1;
10     fail[1] = 0;
11     for(int i = 0,j = -1; word[i]; ++i) {
12         while(j != -1 && word[i] != word[j]) j = fail[j];
13         fail[i+1] = ++j;
14         if(word[i+1] == word[j]) fail[i+1] = fail[j];//使用此句加优化，注释掉不加优化，都是正确的
15     }
16 }
17 int main() {
18     int n,ret;
19     scanf("%d",&n);
20     while(n--) {
21         scanf("%s%s",word,text);
22         getFail();
23         for(int i = ret = 0, j = 0; text[i] ; ++i) {
24             while(j > -1 && word[j] != text[i]) j = fail[j];
25             if(!word[++j]) {ret++;j = fail[j];}
26         }
27         printf("%d
",ret);
28     }
29     return 0;
30 }

View Code