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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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

Sample Output

6-1

题意：找出 b 数组在 a 数组中出现的第一个位置。

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思路：KMP板子题。代码如下：

#include 
   
    using namespace std;const int smaxn = 1000000 + 10, pmaxn = 10000 + 10;int s[smaxn], p[pmaxn];int nextArr[pmaxn];int cases, n, m;namespace kmp1 {
   	void getNext() {
   		nextArr[0] = nextArr[1] = 0;		for (int i = 1; i < m; ++i) {
   			int j = nextArr[i];			while (j && p[i] != p[j]) j = nextArr[j];			nextArr[i + 1] = (p[i] == p[j] ? j + 1 : 0);		}	}	int kmp() {
   		if (n < m) return -1;		getNext();		int j = 0;		for (int i = 0; i < n; ++i) {
   			while (j && s[i] != p[j]) j = nextArr[j];			if (s[i] == p[j]) ++j;			if (j >= m) return i + 2 - j;		}		return -1;	}}namespace kmp2 {
   	void getNext() {
   		nextArr[0] = -1, nextArr[1] = 0;		int pos = 2, cn = 0;		while (pos <= m) {
   			if (p[pos - 1] == p[cn]) nextArr[pos++] = ++cn;			else if (cn > 0) cn = nextArr[cn];			else nextArr[pos++] = cn;		}	}	int kmp() {
   		if (n < m) return -1;		getNext();		int i = 0, j = 0;		while (i < n && j < m) {
   			if (s[i] == p[j]) ++i, ++j;			else if (nextArr[j] != -1) j = nextArr[j];			else ++i;			if (j >= m) return i + 1 - j;		}		return -1;	}}int main() {
   	scanf("%d", &cases);	while (cases--) {
   		scanf("%d%d", &n, &m);		for (int i = 0; i < n; ++i) scanf("%d", &s[i]);		for (int i = 0; i < m; ++i) scanf("%d", &p[i]);		printf("%d\n", kmp1::kmp());	}    return 0;}
   

提交结果如下：

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