hdu1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12912    Accepted Submission(s): 5848

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

#include<iostream>  #include<cstdio>  #include<cstring>  using namespace std;    const int maxn = 1000005;  int a[maxn], b[maxn];  int next1[maxn];  int len1, len2;  void getNext(int c[]){      int n = len2;    next1[0] = -1;      int j = -1;      for(int i = 1; i < n; i++){          while(j!=-1&&c[j+1]!=c[i]){              j = next1[j];          }          if(c[j+1]==c[i]){              j += 1;          }          next1[i] = j;      }  }    int kmp(int str1[], int str2[]){      int j = -1;      getNext(str2);       for(int i = 0; i < len1; i++){          while(j!=-1&&str2[j+1]!=str1[i]){              j = next1[j];          }          if(str2[j+1] == str1[i]){              j += 1;          }           if(j==len2-1){              return i-j+1;        }      }      return -1;  }    int main(){   // freopen("in.txt", "r", stdin);       int n;     cin>>n;     while(n--){     cin>>len1>>len2;     for(int i = 0; i < len1; i++){     scanf("%d",&a[i]); }for(int i = 0; i < len2; i++){scanf("%d",&b[i]);}cout<<kmp(a, b)<<endl; }    return 0;  }