HDU 1711 Number Sequence

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

题意：给定两个数组，输出第一个数组中与第二个数组完全匹配的首个元素的位置，若不能匹配，输出-1

思路：
N最大为1000000，用KMP算法

#include <iostream>using namespace std;int N,M;int a[1000010],b[10005],Next[10005];void getNext()//得到数组b中每个元素的next值{    Next[0]=-1;    int k=-1,j=0;    while(j<M)    {        if(k==-1||b[k]==b[j])        {            k++;            j++;            Next[j]=k;        }        else            k=Next[k];//递归找长度更短的相同前缀后缀    }}int KMP(){    int i=0,j=0;    while(i<N&&j<M)    {        if(j==-1||a[i]==b[j])//若匹配，移动到下一位        {            j++;            i++;        }        else            j=Next[j];//不匹配，i不变，下标j移动到数组b中上一个值为b[j]的位置    }    if(j==M)        return i-j+1;    else        return -1;}int main(){    int T,ans;    cin>>T;    while(T--)    {        cin>>N>>M;        if(M<=N)        {            for(int i=0;i<N;i++)                cin>>a[i];            for(int i=0;i<M;i++)                cin>>b[i];            getNext();            ans=KMP();            cout<<ans<<endl;        }        else//若M>N,不能匹配            cout<<"-1"<<endl;    }}