1711 Number Sequence（kmp）

Number Sequence

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 90   Accepted Submission(s) : 57

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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 

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <string>
 using namespace std;
 int a[];
 int b[];
 int nxt[];
 int n, m;
 int kmp()
 {
     int i, j;
     j = ;
     for (i = ; i <= n; i++)
     {
         while (j&&b[j + ] != a[i]) j = nxt[j];
         if (a[i] == b[j+])
         {
             j++;
         }
         if (j == m)
         {
             return i - m + ;
         }
     }
     return -;
 }
 int main()
 {
     int t;
     cin >> t;
     while (t--)
     {
         cin >> n >> m;
         int i, j;
         for (i = ; i <= n; i++)
         {
             cin >> a[i];
         }
         for (i = ; i <= m; i++)
         {
             cin >> b[i];
         }
         j = ;
         for (i = ; i <= m; i++)
         {
             while (j&&b[j + ] != b[i]) j = nxt[j];
             if (b[j + ] == b[i]) j++;
             nxt[i] = j;
         }
         int f = kmp();
         cout << f << endl;
     }
     return ;
 }

#include <iostream>#include <algorithm>#include <cstring>#include <string>using namespace std;int a[1000005];int b[10005];int nxt[10005];int n, m;int kmp(){int i, j;j = 0;for (i = 1; i <= n; i++){while (j&&b[j + 1] != a[i]) j = nxt[j];if (a[i] == b[j+1]){j++;}if (j == m){return i - m + 1;}}return -1;}int main(){int t;cin >> t;while (t--){cin >> n >> m;int i, j;for (i = 1; i <= n; i++){cin >> a[i];}for (i = 1; i <= m; i++){cin >> b[i];}j = 0;for (i = 2; i <= m; i++){while (j&&b[j + 1] != b[i]) j = nxt[j];if (b[j + 1] == b[i]) j++;nxt[i] = j;}int f = kmp();cout << f << endl;}return 0;}