HDU 1711

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

题目大意：给出两个数字串：文本串和模式串，求模式串在文本串中第一次出现的位置。

解题思路：依然是KMP模板题……

#include<cstdio>
#include<cstring>
#define MAXpat 10000+5
#define MAXstr 1000000+5
int len1,len2;
int Next[MAXpat],str[MAXstr],pat[MAXpat];
void getNext()
{
    int i=0, j=-1;
    Next[0]=-1;
    while(i<len2)
    {
        if(j == -1 || pat[i] == pat[j]) Next[++i]=++j;
        else j=Next[j];
    }
}
int kmp()
{
    getNext();
    int i=0, j=0;
    while(i<len1)
    {
        if(j == -1 || str[i] == pat[j]) i++, j++;
        else j=Next[j];
        if(j == len2) return i-j+1;
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&len1,&len2);
        for(int i=0;i<len1;i++) scanf("%d",&str[i]);
        for(int i=0;i<len2;i++) scanf("%d",&pat[i]);
        printf("%d
",kmp());
    }
}