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最新推荐文章于 2020-07-11 12:42:10 发布

原创最新推荐文章于 2020-07-11 12:42:10 发布 · 207 阅读

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本文介绍了一种使用KMP算法进行高效字符串匹配的方法。通过解析一个具体的问题实例——寻找两个数列间的关系，文章详细阐述了如何利用KMP算法来解决这一问题。包括KMP算法的基本原理、关键步骤如next数组的构建，以及完整的C++实现代码。

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Number Sequence

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.

InputThe 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].
OutputFor 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

用 KMP 来进行字符串匹配

#include<iostream>
using namespace std;
int a[1000005], b[1000005];
int n, m, ne[1000005];
void GetNext(int *p)
{
    int i = 0, j = -1;
    ne[0] = -1;
    while (i < m)
    {
        if (j == -1 || p[i] == p[j]) ne[++i] = ++j;
        else j = ne[j];
    }
}
int Kmp(int *a,int *b)
{
    int i = 0, j = 0;
    while (i < n&&j < m)
    {
        if (j == -1 || a[i] == b[j]) ++i, ++j;
        else j = ne[j];
    }
    if (j == m) return i - j + 1;
    else return -1;
}
int main(){
    int i, N;
    cin>>N;
    while(N--){
        cin>>n>>m;
        for (i = 0; i < n; ++i)
            cin>>a[i];
        for (i = 0; i < m; ++i)
            cin>>b[i];
        GetNext(b);
        cout<<Kmp(a,b)<<endl;
    }
    return 0;
}

有点晕...

传送门：KMP 算法详细讲解