hdu1711 Number Sequence kmp应用

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1711

 

题目：

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

 

 思路：裸的kmp应用，直接用kmp就好了，就当练手

代码：

#include <bits/stdc++.h>
#define PB push_back
#define MP make_pair
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
#define PI acos((double)-1)
#define E exp(double(1))
#define K 1000000+9
int n,m;
int nt[K],a[K],b[K];
void kmp_next(void)
{
    memset(nt,0,sizeof(nt));
    for(int i=1,j=0;i<m;i++)
    {
        while(j&&b[i]!=b[j])j=nt[j-1];
        if(b[i]==b[j])j++;
        nt[i]=j;
    }
}
int kmp(void)
{
    if(m>n)return -1;
    kmp_next();
    for(int i=0,j=0;i<n;i++)
    {
        while(j&&a[i]!=b[j])j=nt[j-1];
        if(a[i]==b[j])j++;
        if(j==m)
            return i-m+2;
    }
    return -1;
}
int main(void)
{
    int t;cin>>t;
    while(t--)
    {
        cin>>n>>m;
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&b[i]);
        printf("%d\n",kmp());
    }

    return 0;
}