hdu1711

Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8078 Accepted Submission(s): 3670

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

#include<stdio.h>  
#include<string.h>  
int next[10005],lena,lenb;  
int a[1000005],b[10005];  
void set_naxt()//子串的next數組  
{  
    int i=0,j=-1;  
    next[0]=-1;  
    while(i<lenb)  
    {  
        if(j==-1||b[i]==b[j])  
        {  
            i++; j++;  
            next[i]=j;  
        }  
        else  
        j=next[j];  
    }  
}  
int kmp()  
{  
    int i=0,j=0;//比較時j＝0  
    set_naxt();  
    while(i<lena)  
    {  
        if(j==-1||a[i]==b[j])  
        {  
            i++;j++;  
        }  
        else  
        j=next[j];//在這裡有可能等於－1，  
  
        if(j==lenb)  
        return i-j+1;  
    }  
    return -1;  
}  
int main()  
{  
    int i,t;  
    scanf("%d",&t);  
    while(t--)  
    {  
        memset(next,0,sizeof(next));  
        scanf("%d%d",&lena,&lenb);  
        for(i=0;i<lena;i++)  
        scanf("%d",&a[i]);  
        for(i=0;i<lenb;i++)  
        scanf("%d",&b[i]);  
        printf("%d\n",kmp());  
    }  
}