ACM-牛喝水

题目描述：牛喝水 

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?

输入

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

输出

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.

样例输入

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

样例输出

3


思路：就像是背包问题或者是开关灯问题，每到一步有放和不放/开不开 两种状态，每种状态做尝试，遍历搜索即可。
备注：之前尝试DFS模拟，但是反转的情况考虑不尽，所以直接用每步判断比较省事。

// 牛喝水.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"


#include <iostream>
using namespace std;

const int MAX = 1000;

int n = 20, ans, flag, arr[MAX];

int check()
{
    for (int i = 0; i < n; i++)
    {
        if (arr[i] == 1)
            return 0;
    }
    return 1;
}

void printa()
{
    for (int i = 0; i < n; i++)
    {
        cout << arr[i] << " ";
    }
    cout << endl;
}

void change(int pos)
{
    arr[pos] = !arr[pos];
    if (pos - 1 >= 0) arr[pos - 1] = !arr[pos - 1];
    if (pos + 1 <= n) arr[pos + 1] = !arr[pos + 1];
}

//也许是反转的情况考虑少了。。。。。
//void DFS(int a[])
//{
//    printa();
//
//    if (is(a) == 1) return;
//
//    for (int i = 0; i < n; i++)
//    {
//        if (a[i] == 1)
//        {
//            //cout << "i:" << i << "\ta[i]:" << a[i] << endl;
//            if ((i+1) < n && a[i + 1] == 1)
//            {
//                a[i] = change(a[i]);
//                a[i + 1] = change(a[i + 1]);
//                if (i + 2 < n) a[i + 2] = change(a[i + 2]);
//            }            
//            else
//            {
//                a[i] = change(a[i]);
//                if (i - 1 >= 0) a[i - 1] = change(a[i - 1]);
//                if (i + 1 < n) a[i + 1] = change(a[i + 1]);
//            }
//            num++;
//            break;
//
//        }
//        
//    }
//    DFS(a);
//}

void DFS(int pos,int sum, int start)
{
    //cout << "pos:" << pos << "\tsum" << sum << "\tstart:" << start << endl;
    if (flag) return;
    if (sum == start) { flag = check(); ans = sum;  return; }
    
    if (n - pos + 1 < start - sum) return;

    change(pos);
    DFS(pos + 1, sum + 1,start);

    change(pos);
    DFS(pos + 1, sum, start);
}


int main()
{
    for (int i = 0; i < n; i++) cin >> arr[i];

    flag = 0;
    ans = -1;
    for (int i = 0; i < n; i++)
    {
        DFS(0, 0, i);
        if (flag)
        {
            cout << ans << endl;
            break;
        }
    }
    if (flag == 0) cout << "20" << endl;

    return 0;
}