hdu1711Number Sequence

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9148 Accepted Submission(s): 4180

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>using namespace std;const int MAXN = 1000005;const int MAXM = 10005;int text[MAXN]; /*文本串*/int next[MAXM]; /*next数组*/int pattern[MAXM];/*模式串*/int m,n,t;/*O(m)的时间求next数组*/void get_next(){    int i=0,j=-1;    next[i]=j;    while(i<m){        if(j==-1 || pattern[i]==pattern[j]){            i++; j++;            next[i]=j;        }else            j = next[j];    }}/*o(n)的时间进行匹配*///void kmp(){//    int i=0, j=0;//    while(i<n && j<m){//        if(j==-1 || text[i]==pattern[j]){//            i++;//            j++;//        }else//            j=next[j];//    }//    if(j>=m) cout<<i-j+1<<endl;//    else cout<<-1<<endl;//}void kmp(){    int j=0;    for(int i=0; i<n; i++){        while(j && text[i]!=pattern[j]) j=next[j];        if(text[i]==pattern[j]) j++;        if(j==m) {            cout<<i-m+2<<endl;            return;        }    }    cout<<-1<<endl;}int main(){    cin>>t;    while(t--){        cin>>n>>m;        for(int i=0; i<n; i++) cin>>text[i];        for(int i=0; i<m; i++) cin>>pattern[i];        get_next();        kmp();    }}

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