hdu-1711 kmp算法学习

看了两篇博客

要求：

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

分析：

其实没啥可分析的，题目大意很好理解，就是给两个大的数组让你匹配，kmp算法上面两篇博客讲的很清楚了，有个细节，kmp算法返回值为数组元素的下标，而该题要求为元素的位置，那么再加一即可，以下为ac代码：

#include <cstdio>
#include <cstring>
using namespace std;
int a[1000005], b[10005];
int next[10005];
int N, M;
void get_next() {
	int k = 0, i = 0;
	next[0] = 0;
	for (i = 1;i < M;i++) {
		while (k > 0 && b[i] != b[k])
			k = next[k - 1];
		if (b[i] == b[k])
			k++;
		next[i] = k;
	}
}
int kmp() {
	int k = 0, i;
	get_next();
	for (i = 0;i < N;i++) {
		while (k > 0 && a[i] != b[k]) { k = next[k - 1]; }
		if (a[i] == b[k])
			k++;
		if (k == M)return i - k + 2;
	}
	return -1;
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--) {
		memset(a, 0, sizeof(a));
		memset(b, 0, sizeof(b));
		memset(next, 0, sizeof(next));
		scanf("%d%d", &N, &M);
		for (int i = 0;i < N;i++)scanf("%d", a + i);
		for (int j = 0;j < M;j++)scanf("%d", b + j);
		printf("%d\n", kmp());
	}
}

来源：CSDN

作者：0cean_star

链接：https://blog.csdn.net/qq_41852483/article/details/103743903

标签

kmp