HDU 1711 Number Sequence

最新推荐文章于 2020-07-12 20:18:12 发布

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最新推荐文章于 2020-07-12 20:18:12 发布

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分类专栏： KMP

本文链接：https://blog.youkuaiyun.com/wust_ZJX/article/details/40743671

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本文介绍了一个基于KMP算法的问题解决方案，该问题是寻找两个数列间首个完全匹配子序列的起始位置。通过详细的代码实现，展示了如何高效地解决此类字符串匹配问题。

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11789 Accepted Submission(s): 5380

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

裸地KMP，字符匹配，求出第一个匹配数字的位置

#include <iostream>
#include<stdio.h>
#include<string>
#include<cstring>
#include<algorithm>
#define N 1000010
using namespace std;

int f[N];
int a[N],b[N];
int n,m;
int main()
{
    int ca;
    while(~scanf("%d",&ca))
    {
        while(ca--)
        {
           scanf("%d %d",&n,&m);
           for(int i=0;i<n;i++)
           scanf("%d",&a[i]);
           for(int i=0;i<m;i++)
           scanf("%d",&b[i]);

            int j=-1;
            f[0]=-1;
            for(int i=1;i<m;i++)
            {
                while(j>=0&&b[j+1]!=b[i])
                   j=f[j];
                   if(b[j+1]==b[i])
                   j++;
                   f[i]=j;
            }
            j=-1;

            int ans=-1;
            for(int i=0;i<n;i++)
            {
                while(j>=0&&b[j+1]!=a[i])
                  j=f[j];
                  if(b[j+1]==a[i])
                  j++;
                  if(j==m-1)
                  {
                      ans=i-m+1+1;
                      break;
                  }
            }
            if(ans==-1)
            {
                printf("-1\n");
            }
            else
            printf("%d\n",ans);
        }
    }
    return 0;
}