HDU 1711 Number Sequence(KMP)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9548    Accepted Submission(s): 4370

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

 

Sample Input

   
   
   
   

    
    
    
    2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6
-1
   
   
   
   

 

题意： 求最先匹配的位置。

思路：第一次做KMP，这题裸题，熟悉一下。

代码：

#include <stdio.h>
#include <string.h>

const int N = 1000005;
const int M = 10005;
int n, m, seq[N], seq1[M], next[M], i, t;


void get_next(int *seq, int m) {
	memset(next, 0, sizeof(next));
	int j = 0;
	for (int i = 2; i <= m; i++) {
		while (j > 0 && seq[i] != seq[j + 1])
			j = next[j];
		if (seq[i] == seq[j + 1]) j++;
		next[i] = j;
	}
}

int kmp(int *seq, int *seq1, int n, int m) {
	int j = 0, ans = 0;
	for (int i = 1; i <= n; i++) {
		while (j > 0 && seq[i] != seq1[j + 1]) j = next[j];
		if (seq[i] == seq1[j + 1]) j++;
		if (j == m) {
			return i - m + 1;
		}
	}
	return -1;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		for (i = 1; i <= n; i++)
			scanf("%d", &seq[i]);
		for (i = 1; i <= m; i++)
			scanf("%d", &seq1[i]);
		get_next(seq1, m);
		printf("%d\n", kmp(seq, seq1, n, m));
	}
	return 0;
}