POJ 3185 The Water Bowls(开关问题）

The Water Bowls

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5941 Accepted: 2344

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]

有20个碗排成一排，有些碗口朝上，有些碗口朝下。每次可以反转其中的一个碗，但是在反转该碗时，该碗左右两边的碗也跟着被反转（如果该碗为边界上的碗，则只有一侧的碗被反转）。求最少需要反转几次，可以使得所有碗口均朝上

这题很坑，，测试数据，，

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

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

可以从俩边遍历取最小值，不然考虑的情况太多容易卡特殊数据，，，

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int dir1[30],dir2[30], f[30];


int main()
{
    while(scanf("%d",&dir1[0])!=EOF)
    {
        memset(f,0,sizeof(f));
        for(int i=1; i<20; i++)
        {
            scanf("%d", &dir1[i]);
        }
        memcpy(dir2,dir1,sizeof(dir1));
        int cnt1=0, cnt2=0, res;
        for(int i=0;i<=19;i++)
        {
            if(dir1[i]==1)
            {
                if(i==19)
                {
                    cnt1=inf;
                    break;
                }
                cnt1++;
                dir1[i+1]=!dir1[i+1];
                dir1[i+2]=!dir1[i+2];
            }
        }
        for(int i=19;i>=0;i--)
        {
            if(dir2[i]==1)
            {
                if(i==0)
                {
                    cnt2=inf;
                    break;
                }
                cnt2++;
                dir2[i-1]=!dir2[i-1];
                dir2[i-2]=!dir2[i-2];
            }
        }
        res=min(cnt1,cnt2);
        printf("%d\n",res);
    }


    return 0;
}