Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 36428 Accepted Submission(s): 15078
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<algorithm> #include<string.h> using namespace std; #define MAXN 1000500 int s[MAXN]; int t[MAXN]; int n,m; int nxt[MAXN]; void getnxt() { int k,j; j=0; k=-1; nxt[0]=-1; while(j<m) { if(k==-1||t[j]==t[k]) { nxt[++j]=++k; } else k=nxt[k]; } } int kmp() { int i=0; int j=0; getnxt(); while(i<n&&j<m) { if(j==-1||s[i]==t[j]) { i++; j++; } else { j=nxt[j]; } } if(j==m) return i-m+1; else return -1; } int main() { ios::sync_with_stdio(0); cin.tie(0); int T; cin>>T; while(T--) { memset(nxt,0,sizeof(nxt)); cin>>n>>m; for(int i=0;i<n;i++) cin>>s[i]; for(int j=0;j<m;j++) cin>>t[j]; cout<<kmp()<<endl; } return 0; }