HDU1711(KMP)

Posted 2020-08-13 Penn000

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22739    Accepted Submission(s): 9727

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

 

 1 //2016.10.7
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 const int inf = 0x3f3f3f3f;
 9 int a[1000005], b[10005], nex[10005];
10 
11 int kmp(int a[], int b[])
12 {
13     int ans = 0;
14     nex[0] = -1;
15     for(int i = 0, fail = -1; b[i] != inf;)//求nex数组， fail为失配指针
16     {
17         if(fail==-1 || b[i] == b[fail])
18         {
19             i++, fail++;
20             nex[i] = fail;
21         }else fail = nex[fail];
22     }
23     int i = 0, j = 0;
24     for(; a[i] != inf; i++, j++)
25     {
26         if(j != -1 && b[j] == inf)return i-j+1;
27         while(j != -1 && a[i] != b[j])j = nex[j];
28     }
29     if(b[j] == inf)return i-j+1;
30     return -1;
31 }
32 
33 int main()
34 {
35     int T, n, m;
36     scanf("%d", &T);
37     while(T--)
38     {
39         scanf("%d%d", &n, &m);
40         for(int i = 0; i < n; i++)scanf("%d", &a[i]);
41         for(int i = 0; i < m; i++)scanf("%d", &b[i]);
42         a[n] = b[m] = inf;//设置截止符号
43         printf("%d\n", kmp(a, b));
44     }
45 
46     return 0;
47 }