hdu1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39044    Accepted Submission(s): 16125

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

 

分析：此题为KMP模板，直接模板即可。

ac代码：

#include <stdio.h>

int a[1000000+50];
int b[10000+50];
int next[10000+50];
int Kmp(int lena,int lenb);
void get_Next(int lenb);

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, m;
        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]);
        get_Next(m);
        printf("%d\n", Kmp(n, m));
    }
    return 0;
}
//KMP算法
int Kmp(int lena,int lenb)
{
    int i=0,j=0;
    while(i<lena&&j<lenb)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
        }
        else
            j=next[j];
    }
    if(j==lenb)
        return i-lenb+1;
    else
        return -1;
}
void get_Next(int lenb)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<lenb-1)
    {
        if(j==-1||b[i]==b[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}