hdu 1711 Number Sequence (KMP)

最新推荐文章于 2019-07-14 13:24:47 发布

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本文介绍了一个关于KMP算法的应用题目，旨在寻找两个数列间匹配的起始位置。通过详细的代码实现，展示了如何利用KMP算法高效解决该问题。

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28879 Accepted Submission(s): 12146

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

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题意：找到第一个匹配的下标。

裸的KMP，直接代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int t[1000005],p[10005],ne[10005];
void makeNext(const int p[],int ne[],int len)
{
    ne[0]=0;
    for(int i=1,k=0;i<len;i++)
    {
        while(k>0&&p[i]!=p[k]) k=ne[k-1];
        if(p[i]==p[k]) k++;
        ne[i]=k;
    }
}
int kmp(const int t[],const int p[],int ne[],int n,int m)
{
    int ans=0;
    makeNext(p,ne,m);
    for(int i=0,k=0;i<n;i++)
    {
        while(k>0&&t[i]!=p[k]) k=ne[k-1];
        if(t[i]==p[k]) k++;
        if(k==m)
        {
            ans=i-m+2;
            return ans;
        }
    }
    return -1;
}
int main()
{
    int c;
    scanf("%d",&c);
    for(int i=1;i<=c;i++)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&t[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&p[i]);
        printf("%d\n",kmp(t,p,ne,n,m));
    }
    return 0;
}