hdu1711 Number Sequence KMP

最新推荐文章于 2021-08-01 08:48:21 发布

cloudy_happy

最新推荐文章于 2021-08-01 08:48:21 发布

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分类专栏： KMP 文章标签： KMP

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本文介绍了一个基于KMP算法的问题实例，详细阐述了如何通过构建next数组来高效匹配两个数列，使得一个数列能作为另一个数列的子序列出现，并提供了完整的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.

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

思路：

KMP板子题

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = (int)1e6 + 10;
int a[10010],b[maxn];
int nxt[10010],n,m;
void init()
{
    int i = 1,j = 0;
    nxt[0] = 0;
    while (i < m)
    {
        if (a[i] == a[j]) nxt[i ++] = ++j;
        else if (!j) i ++;
        else j = nxt[j - 1];
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while (t --)
    {
        scanf("%d %d",&n,&m);
        for (int i = 0;i < n;i ++) scanf("%d",&b[i]);
        for (int i = 0;i < m;i ++) scanf("%d",&a[i]);
        init();
        int ans = -1,i = 0,j = 0;
        while (i < n)
        {
            if (b[i] == a[j])
                i ++,j ++;
            else if (!j) i ++;
            else j = nxt[j - 1];
            if (j == m)
            {
                ans = i - m + 1;
                break;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}