簡單KMP Number Sequence

Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22055 Accepted Submission(s): 9428

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

/*"部分匹配值"就是"前綴"和"後綴"的最長的共有元素的長度。以"ABCDABD"爲例，

　　－　"A"的前綴和後綴都爲空集，共有元素的長度爲0；

　　－　"AB"的前綴爲[A]，後綴爲[B]，共有元素的長度爲0；

　　－　"ABC"的前綴爲[A, AB]，後綴爲[BC, C]，共有元素的長度0；

　　－　"ABCD"的前綴爲[A, AB, ABC]，後綴爲[BCD, CD, D]，共有元素的長度爲0；

　　－　"ABCDA"的前綴爲[A, AB, ABC, ABCD]，後綴爲[BCDA, CDA, DA, A]，共有元素爲"A"，長度爲1；

　　－　"ABCDAB"的前綴爲[A, AB, ABC, ABCD, ABCDA]，後綴爲[BCDAB, CDAB, DAB, AB, B]，共有元素爲"AB"，長度爲2；

　　－　"ABCDABD"的前綴爲[A, AB, ABC, ABCD, ABCDA, ABCDAB]，後綴爲[BCDABD, CDABD, DABD, ABD, BD, D]，共有元素的長度爲0。*/
//next的含義就是這個字符串的前綴和後綴的最大的相同元素
//MAP的精髓所在   可以控制前移的位數  next的回溯十分重要
//理解KMP一定要弄懂next值所代表的含義
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>

using namespace std;
const int MAXX=1000000+10;
int a1[MAXX];
int a2[MAXX];
int nex[MAXX];
int n;
int k;
void Getnext(){
   int i=0,j=-1;
    nex[0]=-1;
    while(i<k)
    {
        if(j==-1||a2[i]==a2[j])
           nex[++i]=++j;
        else
           j=nex[j];

    }
}

void KMP()
{
    int  num=0; int i=0;
    while(i<n){
        if(num==-1||a1[i]==a2[num]){
            num++;i++;
            if(num==k){
                printf("%d\n",i-num+1);
                return ;
            }
        }
        else
            num=nex[num];

    }
    printf("-1\n");
}
int main(){
    int T; scanf("%d",&T);
    while(T--){
        scanf("%d %d",&n,&k);
        for(int i=0;i<n;i++)
            scanf("%d",&a1[i]);
        for(int i=0;i<k;i++)
            scanf("%d",&a2[i]);
        Getnext();
        KMP();
    }
    return 0;
}