Number Sequence ----HDOJ 1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9058    Accepted Submission(s): 4149

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

思路：KMP算法，已经略懂

AC代码：

#include<stdio.h>
int a[1000005], b[10005]; 
int fail[10005];
int n, m, T;  
void getfail()
{
    fail[0] = -1; 
    int i, j; 
    for(i = 1, j = -1; i < m; i ++)
    {
        while(j >= 0 && b[j + 1] != b[i])
        {
            j = fail[j]; 
        }
        if(b[j + 1] == b[i])
            j ++; 
        fail[i] = j; 
    }
    return ; 
}

int kmp()
{
    getfail();
    int i, j;
    for(i = 0, j = 0; i < n;i ++)
    {
        while(j && b[j] != a[i])
        {
            j = fail[j - 1] + 1;
        }
        if(b[j] == a[i])
            j ++;
        if(j == m)
            return i - m + 2;
    }
    return -1;
}

int main(int argc, char const *argv[]) 
{
    scanf("%d", &T);
    int i, j;
    while(T--)
    {
        scanf("%d%d", &n, &m);
        for(i = 0; i < n; i ++)
            scanf("%d", &a[i]);
        for(i = 0; i < m; i ++)
            scanf("%d", &b[i]);
        getfail();
        printf("%d
", kmp());
    }
    return 0;
}