HDU 1711 Number Sequence //简单kmp

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10570    Accepted Submission(s): 4813

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
   
   
   
   

 

Source

HDU 2007-Spring Programming Contest

 

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#include <stdio.h>

int a[10005], b[1000005], next[10005];

void findnext(int m)
{
    int i, j;
    i=0;
    j=-1;
    next[0] = -1;

    while(i<m)
    {
        if(j==-1 || a[i]==a[j])
        {
            i++;
            j++;
            next[i] = j;
        }
        else        
            j = next[j];        
    }
}

int main()
{
    int t;
    int n, m;
    int i, j, flag;

    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for(i=0; i<n; i++)
            scanf("%d", &b[i]);
        for(i=0; i<m; i++)
            scanf("%d", &a[i]);
        findnext(m);

        flag = 0;
        i = 0;
        j = 0;
        while(i<n)
        {
            if(j==-1 || b[i]==a[j])
            {
                i++;
                j++;
                if(j>=m)
                {
                    printf("%d\n", i-j+1);                    
                    flag = 1;
                    break;
                }
            }
            else
                j = next[j];
        }
        if(flag == 0)
            printf("-1\n");
    }

    return 0;
}