[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher A

[kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher

A - Number Sequence

HDU - 1711

题目链接：https://vjudge.net/contest/70325#problem/A

题目：

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.

InputThe 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].
OutputFor 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
题意：给你两个序列a,b，找出b序列在a中出现的位置，利用kmp算法

//
// Created by hanyu on 2019/8/13.
//

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#include<math.h>
#include<map>
using namespace std;
typedef long long ll;
const int maxn=1e6+7;
#define MAX 0x3f3f3f3f
int a[maxn],b[maxn];
int nextt[maxn];
int n,m;
void getnext()
{
    int i=0,j=-1;
    nextt[0]=-1;
    while(i<m) {
        if (j == -1 || b[i] == b[j]) {
            i++, j++;
            if (b[i] != b[j])
                nextt[i] = j;
            else
                nextt[i] = nextt[j];
        } else
            j = nextt[j];
    }
}
int kmp()
{
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++,j++;
        } else
            j=nextt[j];
    }
    if(j==m)
        return i-j+1;
    return -1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int j=0;j<m;j++)
            scanf("%d",&b[j]);
        getnext();
        if(n<m)
            printf("-1
");
        else
            printf("%d
",kmp());
    }
    return 0;
}