HDU 1711 Number Sequence（KMP）

最新推荐文章于 2021-03-01 20:47:37 发布

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最新推荐文章于 2021-03-01 20:47:37 发布

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分类专栏： KMP 文章标签： KMP

本文链接：https://blog.csdn.net/j2_o2/article/details/81485809

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本文详细介绍了一种高效的字符串匹配算法——KMP算法，并通过一个具体的编程实例来解释如何使用该算法解决特定的问题。文章首先介绍了算法的基本原理，接着通过示例展示了如何实现KMP算法，并给出了完整的代码实现。

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Problem Description

Given two sequences of numbers : a1, a[2], …… , a[N], and b1, b[2], …… , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b1, 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 a1, a[2], …… , a[N]. The third line contains M integers which indicate b1, 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板子题，学习KMP点我,了解算法思想，这里好像没有讲为什么回溯到j= nxt[j]。
给组回溯了两次的样例手推一下，用心感受(ಥ_ಥ)
需要匹配的串：abcababcaab
nxt数组：-1 0 0 0 1 2 1 2 3 4 1 （此处首位为-1

代码：

#include<stdio.h>

int n, m;
int nxt[10005], s1[1000005], s2[10005];

void getnxt()
{
    nxt[0] = -1;
    int i = 0, j = -1;
    while(i<m)
    {
        if(j == -1 || s2[i] == s2[j]) nxt[++i] = ++j;
        else j = nxt[j];
    }
}

int kmp()
{
    int i = 0, j = 0;
    while(i < n)
    {
        if(j == -1 || s1[i] == s2[j]) ++i, ++j;
        else j = nxt[j];
        if(j == m) return i+1-m; // 匹配成功
    }
    return -1;
}

int main()
{
    int __;
    for(scanf("%d",&__); __; --__)
    {
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; ++i) scanf("%d",&s1[i]);
        for(int i = 0; i < m; ++i) scanf("%d",&s2[i]);
        getnxt();
//        for(int i = 0; i < m; ++i) printf("%d ",nxt[i]);printf("\n");
        printf("%d\n",kmp());
    }
    return 0;
}