題意：

給20個水碗。朝上為‘0’或朝下為‘1’，每次操作使三個碗翻轉(zhuǎn)，問使所有20個水碗都朝上，至少翻多少次？

題目：

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]

分析：

這個題直接暴力過得，直接正反遍歷，找最小值，還有一種構(gòu)造增廣矩陣類似于根據(jù)開關(guān)的關(guān)系構(gòu)造有向圖的鄰接矩陣；構(gòu)造增廣矩陣，高斯消元，枚舉自由元（二進制枚舉狀態(tài)），尋找最小值的方法，因為時間關(guān)系就沒有去鉆研，在這提供下思路。

AC代碼：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int a[30],b[30]; int ans,num; int main(){for(int i=1;i<=20;i++){scanf("%d",&a[i]);b[i]=a[i];}num=ans=0;for(int i=1;i<=20;i++){if(a[i]==1){ans++;a[i+1]=!a[i+1];a[i+2]=!a[i+2];}}for(int i=20;i>=1;i--){if(b[i]==1){num++;b[i-1]=!b[i-1];b[i-2]=!b[i-2];}}ans=min(ans,num);printf("%d\n",ans);return 0; }

總結(jié)

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