hdu_1711Number Sequence(kmp)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20445    Accepted Submission(s): 8756

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

赤裸裸的kmp

//kmp
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[1000005],b[10005];
int Next[10005];
int n,m;
int kmp()
{
    int i,j;
    j = 0;
    int tm = Next[0] = -1;
    //ÇóNextÊý×é
    while(j<m-1){
        if(tm<0||b[j]==b[tm])
            Next[++j] = ++tm;
        else tm = Next[tm];
    }
    //Æ¥Åä
    for( i = j = 0; i < n&&j < m; ){
        if(j<0||a[i]==b[j])i++,j++;
        else j = Next[j];
    }
    if(j<m) return -1;
    return i-j;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        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]);
        int ans = kmp();
        if(ans!=-1)
        printf("%d
",ans+1);
        else puts("-1");
    }
    return 0;
}