i+1 cool pro flag 關燈問題 rom arr AI integer

題目描述：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?

輸入

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

輸出

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.

樣例輸入

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

樣例輸出

3


思路：就像是背包問題或者是開關燈問題，每到一步有放和不放/開不開 兩種狀態，每種狀態做嘗試，遍歷搜索即可。
備註：之前嘗試DFS模擬，但是反轉的情況考慮不盡，所以直接用每步判斷比較省事。

// 牛喝水.cpp : 定義控制臺應用程序的入口點。
//

#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];
}

//也許是反轉的情況考慮少了。。。。。
//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;
}

ACM-牛喝水