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最新推荐文章于 2022-04-07 18:08:23 发布

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最新推荐文章于 2022-04-07 18:08:23 发布

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分类专栏：数据结构

本文链接：https://blog.youkuaiyun.com/wahaha1_/article/details/8456556

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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

KMP()算法求next[ ]值的确不好理解，下面的三个连接或许可以帮到你：

http://blog.youkuaiyun.com/ts173383201/article/details/7850120

http://www.cnblogs.com/dolphin0520/archive/2011/08/24/2151846.html

http://www.56.com/u59/v_NjAwMzA0ODA.html这是严老师的一段视频,看完之后豁然开朗。

#include<stdio.h>
#include<string.h>
#define max 1000010
int a[max],b[10010];
int next[10010];
int aLen,bLen;
void get_next()
{
    int j,k;
    j=0;
    k=-1;
    next[0]=-1;
    while(j<bLen)
    {
        if(k==-1||b[j]==b[k])
          next[++j]=++k;
        else
          k=next[k];
    }
}
void KMP()
{
    int i,j=0;
    int flag=0;
    if(aLen<bLen)
    {
        printf("-1\n");
        return;
    }
    for(i=0;i<aLen;)//这里i++错误，刚开始思考错了
    {

        if(a[i]==b[j])
        {
            if(j==bLen-1)
            {
                 printf("%d\n",i-bLen+2);
                 flag=1;
                 break;
            }
          j++;i++;
        }
        else
        {
            j=next[j];
            if(j==-1)
            {
                i++;
                j=0;
            }
        }
    }
    if(flag==0)
    printf("-1\n");
}
int main()
{
    int T;
    int i,j;
    scanf("%d",&T);
    while(T--)
    {
        int k=0;
        scanf("%d%d",&aLen,&bLen);
        for(i=0; i<aLen; i++)
            scanf("%d",&a[i]);
        for(j=0; j<bLen; j++)
        {
              scanf("%d",&b[j]);
        }
        get_next();
        KMP();
    }
    return 0;
}