HDU - 1686 Oulipo KMP匹配运用
id=25191" target="_blank" style="color:blue; text-decoration:none">HDU - 1686
Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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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
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
/*
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;
}
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