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C Solve LIS through recursion

Exuberant Elephant answered on April 29, 2022 Popularity 1/10 Helpfulness 1/10

answer C Solve LIS through recursion

C Solve LIS through recursion

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#include<iostream>
using namespace std;
int minStep,n,*arr,*record,*lis,index,recordMax,lisCount;
/*
	1.minStep :存放"只"遍历一次指定数组,得到的LIS的长度.比如:
		*arr={4,5,1,2,3};  遍历该数组过后,minStep=2,即为{4,5}
		两个元素的长度.具体请看getMinStep方法.
	2.*arr : 存放输入的,或者随机产生的一组数;
	3.*record : 存放遍历过程中符合条件的subsequence.
	4.index :  record数组的元素个数
	5.recordMax : record数组中最大的元素
	6.lisCount :  LIS个数. 
	
	整体思想:	
			从第一个数开始往后遍历,每遇到一个数,若它大于之前的数,
			则它可以要,也可以不要. 若要了,为了不影响之后的判断,
			要完之后还得还回来. 
*/
void copy()
{
	for(int i=0;i<lisCount;i++)
		lis[i]=record[i];
}
void dfs(int step)
{
	if(step>=n)
	{
		if(index>lisCount)
		{
			lisCount=index;
			copy();
		}
		return;
	}
	if(n-step+1<minStep-index)
		return;
	else if(recordMax<arr[step+1])//若符合条件,可以放入record 
	{
		int temp=recordMax;//放入record 
		recordMax=arr[step+1];
		record[index]=arr[step+1];
		index++;
		dfs(step+1);
		index--;//取出,以便递归. 
		record[index]=0;
		recordMax=temp;
	}
	dfs(step+1);//不放入record 
}
int getMinStep()//遍历一次数组,得到一个 IS (不是LIS) 用于剪枝. 
{
	int i,count=1,temp=arr[1];
	for(i=2;i<=n;i++)
		if(temp<arr[i])
		{
			temp=arr[i];
			count++;
		}
	return count;
}
void init()//initialzation 
{
	index=0;
	recordMax=0;
	lisCount=0;
		
	arr=new int[n+1];
	record=new int[n+1]();
	lis=new int[n+1]();
		
	arr[0]=0;
	for(int i=1;i<=n;i++)
		cin>>arr[i];
}
void show(int max)
{
	cout<<"LIS length: "<<max<<endl;
	for(int i=0;i<max;i++)
		cout<<lis[i]<<" ";
	cout<<endl;
}
void deleteArray()
{
	delete lis;
	delete record;
	delete arr;
}
 
int main()
{	
	cin>>n;
	
	init();
		
	minStep=getMinStep();
		
	if(minStep==n)//整个数组都是递增的,直接输出. 
	{
		show(n);
		return 0;
	}
		
	dfs(0);
		
	show(lisCount);
		
	deleteArray();
	
	return 0;
}

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Source: blog.csdn.net

Tags: c lis recursion whatever

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Contributed on Apr 29 2022

Exuberant Elephant

0 Answers Avg Quality 2/10

C Solve LIS through recursion

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