HDU 1711 Number Sequence_java hdu1711-数字序列问题-优快云博客

本文链接：https://blog.youkuaiyun.com/wangcong9614/article/details/47101047

本文介绍了一种使用裸KMP算法解决数字序列匹配问题的方法，旨在寻找目标子序列在长序列中的起始位置。通过具体实例展示了算法实现过程，并提供了完整的代码示例。

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裸KMP，求给定的数字序列中是否有给出的子串，即模式串。如果有的话，输出与模式串匹配上的首字母下标。

如果对KMP没有理解的话：附教程地址：KMP及KMP next数组讲解

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14806 Accepted Submission(s): 6496

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

题目链接：HDU1711 Number Sequence

AC代码：

#include<stdio.h>
#include<string.h>
#define N 1000005
#define M 10005
int a[N],b[M];
int next[M];
int n,m;
void setNext()
{
    int i,j;
    i=0;
    j=-1;
    next[i]=j;
    while(i<m)
    {
        if(j==-1||b[i]==b[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
    return ;
}

int KMP()
{
    int i,j;
    i=j=0;
    setNext();
    while(i<n)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
            if(j==m)
                return i-m+1;
        }
        else
            j=next[j];
    }
    return -1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        int i;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(i=0;i<m;i++)
            scanf("%d",&b[i]);
        int ans=KMP();
        printf("%d\n",ans);
    }
    return 0;
}