HDU_1711_初识KMP算法

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22568    Accepted Submission(s): 9639

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

 

 

long long ago:KMP算法还需多理解。

 

2017.2.27 15:09 第一次复习KMP，看着代码什么也想不起来，然后看了两篇博文。

　　1.《KMP算法》  2.《KMP算法的前缀next数组最通俗的解释，如果看不懂我也没辙了》

　　　　写写这次复习的体会，求next数组，我的体会是求模式串str和自己本身的对称性，也如第一篇博文中讲的那样，next[i]即为

　　模式串的子串str[1--i]的所有前缀和所有后缀的最长的共有元素的长度(此处前缀后缀的定义可以在博文1中看到)。

　　　　得到next数组后，即可进行KMP算法，匹配时，若在当前位置j匹配失败，那么就将模式串str右移next[j]位，而不是每次右

　　移一位这就是。

　　多写博客，方便复习。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
using namespace std;

int n,m;
int N[1000005],M[10005],Pi[10005];

void preFix()   //构造next数组（寻找模式串的对称性）
{
    memset(Pi,0,sizeof(Pi));
    int k=0;
    for(int q=2;q<=m;q++)
    {
        while(k>0&&M[k+1]!=M[q]) //不相等，则向前递推，对称性减小
            k=Pi[k];
        if(M[k+1]==M[q])   //若相等，对称性继续扩大
            k++;
        Pi[q]=k;
    }
}

int KMP()
{
    preFix();
    int q=0;
    for(int i=1;i<=n;i++)
    {
        while(q>0&&M[q+1]!=N[i])
            q=Pi[q];
        if(M[q+1]==N[i])
            q++;
        if(q==m)
            return i-m+1;
    }
    return -1;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&N[i]);
        for(int j=1;j<=m;j++)
            scanf("%d",&M[j]);
        int ans=KMP();
        printf("%d
",ans);

    }
    return 0;
}