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HDU1711(KMP)

2024-08-15 22:06:43   217人阅读

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 }

 

HDU1711(KMP)


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