Poj 3185（开关问题）

The Water Bowls

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3790 Accepted: 1483

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]

Source

USACO 2006 January Bronze

一道简单的开关问题，因为是一个序列，所以第一次翻转的位置一定是最左端右面的那个，或者是最右端左面的那个。就是从左和从右两边模拟，答案取较小值。

#include<cstdio>#include<iostream>#include<algorithm>#include<vector>#include<cstring>#include<queue>#include<stack>using namespace std;const int maxn = 20 + 5;const int INF = 2000000000;typedef  pair<int, int> P;typedef long long LL;int a[maxn];int tem[maxn];int main(){    while(scanf("%d",&a[0]) != EOF){        for(int i = 1;i < 20;i++){            scanf("%d",&a[i]);        }        for(int i = 0;i < 20;i++) tem[i] = a[i];        int cnt1 = 0;        for(int i = 0;i < 20;i++){            if(a[i] == 1){                if(i == 19){                    cnt1 = INF;                    break;                }                cnt1++;                a[i+1] = !a[i+1];                a[i+2] = !a[i+2];            }        }        int cnt2 = 0;        for(int i = 19;i >= 0;i--){            if(tem[i] == 1){                if(i == 0){                    cnt2 = INF;                    break;                }                cnt2++;                tem[i-1] = !tem[i-1];                tem[i-2] = !tem[i-2];            }        }        printf("%d\n",min(cnt1,cnt2));    }}