Codeforces Round #612 (Div. 2) C. Garland

最新推荐文章于 2020-11-20 20:00:51 发布

falconfan

最新推荐文章于 2020-11-20 20:00:51 发布

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文章标签： c++ 动态规划

本文链接：https://blog.youkuaiyun.com/weixin_45320798/article/details/104037216

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原题：
Vadim loves decorating the Christmas tree, so he got a beautiful garland as a present. It consists of n light bulbs in a single row. Each bulb has a number from 1 to n (in arbitrary order), such that all the numbers are distinct. While Vadim was solving problems, his home Carp removed some light bulbs from the garland. Now Vadim wants to put them back on.
Vadim wants to put all bulb back on the garland. Vadim defines complexity of a garland to be the number of pairs of adjacent bulbs with numbers with different parity (remainder of the division by 222). For example, the complexity of 1 4 2 3 5 is 2 and the complexity of 1 3 5 7 6 4 2 is 111.

No one likes complexity, so Vadim wants to minimize the number of such pairs. Find the way to put all bulbs back on the garland, such that the complexity is as small as possible.

Input

The first line contains a single integer n(1≤n≤100)n (1≤n≤100)n(1≤n≤100) — the number of light bulbs on the garland.

The second line contains n integers p1,p2,…,pn(0≤pi≤n)p1, p2, …, pn (0≤pi≤n)p1,p2,…,pn(0≤pi≤n) — the number on the $ i-th $ bulb, or 000 if it was removed.

Output

Output a single number — the minimum complexity of the garland.

思想：
这道题主要应用分类讨论+动态规划的思想
对于当前已经确定类型的数，只需将其目前状态用前一状态表示即可
而未确定类型的话，则需讨论其为奇或是为偶，并取其小值

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
const int N=105;
int f[N][N][2],a[N];//当前数，可用奇数，当前数奇偶性 0偶1奇 有多少奇偶性相反的组合 
int main(){
	int n,used_odd=0,used_even=0,has_odd,has_even,even,odd,i,j;
	cin>>n;
	even=n/2;odd=n-even;
	for(i=1;i<=n;i++)
	{
	    cin>>a[i];
	    if(a[i]%2==1)
	        used_odd++;
	    else
	        if(a[i]!=0)
	            used_even++;
	}
	memset(f,0x3f3f3f3f,sizeof(f));
	has_even=even-used_even;
	has_odd=odd-used_odd;
	if(a[1]==0)//第一个就需要选择 
	    f[1][has_odd][0]=f[1][has_odd-1][1]=0;
	else
	    if(a[1]%2==1)
	        f[1][has_odd][1]=0;
	    else
	        f[1][has_odd][0]=0;
	for(i=2;i<=n;i++)
	{
		if(a[i]%2==1)
		    for(j=0;j<=has_odd;j++)
		        f[i][j][1]=min(f[i-1][j][1],f[i-1][j][0]+1);
		else
		    if(a[i]!=0)
		        for(j=0;j<=has_odd;j++)
		            f[i][j][0]=min(f[i-1][j][1]+1,f[i-1][j][0]);
		    else
			    for(j=0;j<=has_odd;j++){
			    	f[i][j][0]=min(f[i-1][j][1]+1,f[i-1][j][0]);
		            f[i][j][1]=min(f[i-1][j+1][1],f[i-1][j+1][0]+1);
				}
	}
	cout<<min(f[n][0][0],f[n][0][1]);
	return 0;
}