Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
Description
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
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
Sample Input
Sample Output
/* Problem : 1686 ( Oulipo ) Judge Status : Accepted RunId : 14384097 Language : G++ Author : 24862486 Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta */ #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; const int MAXN = 10000 + 5; const int MAXM = 1000000 + 5; char W[MAXN], T[MAXM]; int n, nexts[MAXN]; void Get_Next(int M) { int i = 0,j = -1; nexts[0] = -1; while(i < M) { if(j == -1 || W[i] == W[j]) { i ++; j ++; if(W[i] == W[j]) { nexts[i] = nexts[j]; } else { nexts[i] = j; } } else { j = nexts[j]; } } } int Get_KMP(int N,int M) { int i = 0,j = 0 ,num = 0; while(i < N) { if(j == -1 || W[j] == T[i]) { i ++; j ++; } else { j = nexts[j]; } if(j == M) { num ++; j = nexts[j];//直接从此处匹配模式串的地方进行匹配。防止超时 } } return num; } int main() { scanf("%d", &n); while(n --) { scanf("%s%s", W, T); int t1 = strlen(W), t2 = strlen(T); Get_Next(t1); printf("%d\n", Get_KMP(t2, t1)); } return 0; }