HDU1711 Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3852 Accepted Submission(s): 1781

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：其实不必解释。

解题思路：最裸的KMP。

#include <stdio.h>
#include <string.h>
#define N 1000005
#define M 10005

int a[N],b[M];
int n,m;
int p[M];

void re(void)
{
    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 run(void)
{
    memset(p,-1,sizeof(p));
    for(int i=1;i<m;i++)
    {
        int k=p[i-1];
        while(1)
        {
            if(b[i]==b[k+1])
            {
                p[i]=k+1;
                break;
            }
            if(k==-1)break;
            k=p[k];
        }
    }
    int j=-1;
    for(int i=0;i<n;i++)
    {
        while(1)
        {
            if(a[i]==b[j+1])
            {
                j++;
                if(j+1==m)
                {
                    printf("%d\n",i-j+1);
                    return ;
                }
                break;
            }
            if(j==-1)break;
            j=p[j];
        }
    }
    puts("-1");
}

int main()
{
    freopen("/home/fatedayt/in","r",stdin);
    int ncase;
    scanf("%d",&ncase);
    while(ncase--)
    {
        re();
        run();
    }
    return 0;
}