Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24116 Accepted Submission(s): 10232
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 #include<iostream> #include<cstring> using namespace std; int a[1000010],b[1000010],l1,l2; int Case,ne[1000010]; void get_ne() { int j=0,k=-1; ne[0]=-1; while(j<l2) { if(k==-1||b[j]==b[k]) ne[++j]=++k; else k=ne[k]; } } void kmp() { int i=0,j=0; get_ne(); while(i<l1) { if(j==-1||a[i]==b[j])i++,j++; else j=ne[j]; if(j==l2) {cout<<i-l2+1<<endl;return;} } cout<<-1<<endl; return; } int main() { cin>>Case; while(Case--) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); cin>>l1>>l2; for(int i=0;i<l1;i++)cin>>a[i]; for(int j=0;j<l2;j++)cin>>b[j]; kmp(); } }
