【KMP】Number Sequence

KMP算法

KMP的基处题目，数字数组的KMP算法应用。

主要是next[]数组的构造，next[]存储的是字符的当前字串，与子串前字符匹配的字符数。

移动位数 = 已匹配的字符数 - 对应的部分匹配值

 

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

#include<stdio.h>
int a[1000001],b[10001],next[10001];
void getnext(int m){
    int i=1,j=0;
    next[1]=0;
    while(i<m){
        if(j==0||b[i]==b[j]){
            i++; j++; next[i]=j;
        }
        else j=next[j];
    }
}

void getk(int n,int m){
    int i=1,j=1;
    while(i<=n&&j<=m){
        if(j==0||a[i]==b[j]){i++; j++;}
        else j=next[j];
    }
    if(j>m) printf("%d\n",i-m);
    else printf("-1\n");
}

int main()
{
    int t,n,m,i,j;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++) scanf("%d",&a[i]);
        for(i=1;i<=m;i++) scanf("%d",&b[i]);
        getnext(m);
        getk(n,m);
    }
    return 0;
}