A - Number Sequence (KMP)

最新推荐文章于 2020-05-14 16:19:20 发布

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本文介绍了一种利用KMP算法解决特定数字序列匹配问题的方法，通过寻找两个数列之间的对应关系来找到最小起始位置。文章提供了一个完整的C++实现案例，并通过样例输入输出展示了算法的有效性。

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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

kmp不止可以用在字符串，还可以用在数字上……（笑着哭）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int t[1000005], p[10005];
int NEXT[10005];
int n, m;
void getnext(int p[])
{
    int i = 0, j = -1;
    NEXT[0] = -1;
    while(i<m)
    {
        if(j==-1||p[i]==p[j])
        {
            i++;
            j++;
            NEXT[i] = j;
        }
        else
            j = NEXT[j];
    }
}
int kmp(int t[], int p[])
{
    int i = 0, j = 0;
    while(i<n&&j<m)
    {
        if(j==-1||t[i]==p[j])
            i++, j++;
        else
            j = NEXT[j];
    }
    if(j>=m)
        return i - m + 1;
    else
        return -1;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        for(int i=0;i<n;i++)
            cin>>t[i];
        for(int i=0;i<m;i++)
            cin>>p[i];
        if(n<m)
            printf("-1\n");
        else
        {
        getnext(p);
        printf("%d\n", kmp(t, p));
        }
    }
    return 0;
}