POJ3185 The Water Bowls 反转（开关）

Description

The cows have a line of 20 water bowls from which they drink. The bowls can be either right-side-up (properly oriented to serve refreshing cool water) or upside-down (a position which holds no water). They want all 20 water bowls to be right-side-up and thus
use their wide snouts to flip bowls. 



Their snouts, though, are so wide that they flip not only one bowl but also the bowls on either side of that bowl (a total of three or -- in the case of either end bowl -- two bowls). 



Given the initial state of the bowls (1=undrinkable, 0=drinkable -- it even looks like a bowl), what is the minimum number of bowl flips necessary to turn all the bowls right-side-up?

Input

Line 1: A single line with 20 space-separated integers

Output

Line 1: The minimum number of bowl flips necessary to flip all the bowls right-side-up (i.e., to 0). For the inputs given, it will always be possible to find some combination of flips that will manipulate the bowls to 20 0's.

Sample Input

0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0

Sample Output

3

Hint

Explanation of the sample: 



Flip bowls 4, 9, and 11 to make them all drinkable: 

0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [initial state] 

0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 4] 

0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 9] 

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [after flipping bowl 11]

        这题属于反转（开关）类型的题，典型的特征就是：

1，反转的先后顺序是不重要的。

2。主动对一个开关进行2次或2次以上的反转是多余的。

        这题。假设条件是每次必须反转3个碗的话，那么就非常easy，先考虑最左端的碗，假设碗朝下，那么这个碗必须反转，同一时候带动后面两个碗一起反转，这种话问题的规模就降低了一个，然后反复此方法推断。

        可是条件是在两端能够出现同一时候仅仅反转两个碗的情况，这时候仅仅要先枚举一下两端反转两个碗的全部情况，然后就能够把它当成每次必须反转3个碗进行处理就能够了。

#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>
#include <queue>
#include <list>
#include <algorithm>
#include <stack>
#include <map>

using namespace std;

int main()
{
#ifdef _DEBUG
	freopen("e:\\in.txt", "r", stdin);
#endif
	int side[21];
	int side1[21];
	for (int i = 0; i < 20; i ++)
	{
		scanf("%d", &side[i]);
	}
	int minCount = 50;
	for (int i = 0; i < 4;i++)
	{
		int count1 = 0;
		memcpy(side1, side, sizeof(side));
		if (i == 1)
		{
			side1[0]++;
			side1[1]++;
			count1++;
		}
		else if (i == 2)
		{
			side1[18]++;
			side1[19]++;
			count1++;
		}
		else if (i == 3)
		{
			side1[0]++;
			side1[1]++;
			side1[18]++;
			side1[19]++;
			count1++;
			count1++;
		}
		for (int i = 0; i <= 17; i++)
		{
			if (side1[i] & 1)
			{
				side1[i] ++;
				side1[i + 1]++;
				side1[i + 2]++;
				count1++;
			}
		}
		if (!(side1[1] & 1 || side1[18] & 1 || side1[19] & 1))
		{
			if (minCount > count1)
			{
				minCount = count1;
			}
		}
	}
	printf("%d\n", minCount);
	return 1;
}