Time Limit: 5000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

Status

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

Source


Hmm………..KMP模板题,char改成int就好。
AC代码:
#include <cstdio>
using namespace std;
int a[1000010], b[10010];
int n, m;
void input()
{
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
		scanf("%d", &a[i]);
	for (int i = 0; i < m; i++)
		scanf("%d", &b[i]);
}
void kmp_pre(int x[], int m, int next[])
{
	int i, j;
	j = next[0] = -1;
	i = 0;
	while (i < m)
	{
		while (-1 != j && x[i] != x[j])
			j = next[j];
		next[++i] = ++j;
	}
}
int next[10010];
int kmp(int x[], int m, int y[], int n)
{
	int i, j;
	kmp_pre(x, m, next);
	i = j = 0;
	while (i < n)
	{
		while (-1 != j && y[i] != x[j])
			j = next[j];
		i++, j++;
		if (j == m)
			return i - j + 1;
	}
	return -1;
}
void solve()
{
	printf("%d\n", kmp(b, m, a, n));
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		input();
		solve();
	}
	return 0;
}

 

分类: ACM/ICPC

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