随笔分类 (148)

随笔档案 (148)

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

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]

Source

USACO 2006 January Bronze

解题报告：这道题是昨天的省赛选拔三的题，当时我没有负责这道题，今天看了队友写的代码，原来是搜索题！就是通过翻转全部变成0，经过的次数。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX = 21;
int bowl[MAX], step, flag;
bool Judge()
{
    for (int i = 0; i < 20; ++i)
    {
        if (bowl[i])//不全部为0时
        {
            return false;
        }
    }
    return true;
}
void Change(int i)
{
    bowl[i] = !bowl[i];
    if (i > 0)//边界处理
    {
        bowl[i - 1] = !bowl[i - 1];
    }
    if (i < 19)//边界处理
    {
        bowl[i + 1] = !bowl[i + 1];
    }
}
void DFS(int i, int num)
{
    if (step == num)
    {
        flag = Judge();
        return;
    }
    if (i >= 20 || flag)
    {
        return;
    }
    Change(i);//翻转
    DFS(i + 1, num + 1);
    Change(i);//再翻转回来
    DFS(i + 1, num);
}
int main()
{
    int i, ans;
    for (i = 0; i < 20; ++i)
    {
        scanf("%d", &bowl[i]);
    }
    for (step = 0; step < 20; ++step)
    {
        flag = 0;
        DFS(0, 0);
        if (flag)
        {
            ans = step;
            break;
        }
    }
    printf("%d\n", ans);
    return 0;
}