[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 1Sample 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\n");
else
printf("%d\n",kmp());
}
return 0;
}