hdu-1711 kmp 附板子 ʕ •ᴥ•ʔ

最新推荐文章于 2019-07-30 18:53:30 发布

henucm

最新推荐文章于 2019-07-30 18:53:30 发布

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本文链接：https://blog.youkuaiyun.com/henucm/article/details/81870401

本文详细介绍了一种高效的字符串匹配算法——KMP算法，并提供了完整的实现代码。通过对KMP算法原理及其实现过程的深入剖析，帮助读者理解如何在给定的字符串中快速查找子串。

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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

-1

题意：给你一个长度为n的字串一个长度为m的模板串

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#define ll long long
using namespace std;
int a[1000010],b[1000010];
int nexxt[1000010];
int n,m;
void getnext()
{
	int j=-1,i=0;
	nexxt[0]=-1;
	while(i<m)
	{
		if(j==-1||b[i]==b[j])
		{   
			i++;
			j++;
			nexxt[i]=j;
		}
		else
		j=nexxt[j];
	}
	
}
int kmp()
{
	getnext();
	int j=0,i=0;
	while(i<n)
	{
		if(j==-1||a[i]==b[j])
		i++,j++;
		else
		{
			j=nexxt[j];
		}		
		if(j==m)
		{
			return i-m+1;
		}
	}
	return -1;
	
}
int main()
{
	ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--)
	{
		cin>>n>>m;
		for(int i=0;i<n;i++)
		cin>>a[i];
		for(int i=0;i<m;i++)
		{
			cin>>b[i];
		}
		cout<<kmp()<<endl;
	}
	return 0;
}

模板：

void getnext()
{
	int j=-1,i=0;
	nexxt[0]=-1;
	while(i<m)
	{
		if(j==-1||b[i]==b[j])
		{   
			i++;
			j++;
			nexxt[i]=j;
		}
		else
		j=nexxt[j];
	}
	
}
int kmp()
{
	getnext();
	int j=0,i=0;
	while(i<n)
	{
		if(j==-1||a[i]==b[j])
		i++,j++;
		else
		{
			j=nexxt[j];
		}		
		if(j==m)
		{
			//返回你所需要的东西
		}
	}
	return ;//返回你所需要的东西
	
}