B - The Pilots Brothers' refrigerator

B - The Pilots Brothers’ refrigerator

文章目录

B - The Pilots Brothers' refrigerator
题目描述：
Input：
Output：
Sample Input：
题目大意：
思路分析：
Sample Input：
Sample Output：
代码块：

题目描述：

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.

There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input：

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output：

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

Sample Input：

题目大意：

就是有一个特别大的冰箱，有16个把手，每个把手有2种状态，+与-，如果全为-的时候，就可以打开冰箱。4v4中任何一个地方，改变一个地方，就会改变当前位置与所在到行与列的状态。输出最少次数打开冰箱，以及路径。

思路分析：

这道题和flip game 的思路差不多，只是多了一个数组记录状态就行了，也是通过一个一个去搜，就行了，可以用BFS找出最短路径去写，也可以用DFS去压栈，在输出即可，不过自己还是要发一下牢骚，这道题因为错将==写成=导致不知道哪里出问题，卡了整整3个小时，心态炸了。

Sample Input：

Sample Output：

代码块：

 #include<stdio.h>char c1[4][4];//把手状态int x2[1000][16];//记录路径x轴int y2[1000][16];//记录路径y轴int min=22,n1=0,n2;//min只要大于16就行了，随便设。void lemone3(int x,int y,int num);//num记录步数int len()//判断是否全为'-'号{int b,c;for(b=0;b<4;b++){for(c=0;c<4;c++){if(c1[b][c]=='+'){return 0;}}}return 1;}void Lemonscanf()//记录冰箱把手状态{int a,b;for(b=0;b<4;b++){gets(c1[b]);} }void LemonDFS(int x,int y,int num)//深搜开始，num记录步数{int c,x1,y1;if(x>=4)//如果x>=4,说明已经翻转完了{if(len()){if(num<min)//比较步数{min=num;for(c=0;c<min;c++)//这一步类似走迷宫，上一次//步数的路径复制给下一次步数的路径。{y2[c][0]=x2[c][0];y2[c][1]=x2[c][1];}}}return;}//x1=x+(y+1)/4;y1=(y+1)%4;//这2步主要是类似在4V4棋盘中一个一个的翻。lemone3(x,y,num);x2[num][0]=x;//记录路径x轴x2[num][1]=y;//记录路径y轴LemonDFS(x1,y1,num+1);// 这一步是退到上一步，回到上一个节点。lemone3(x,y,num);LemonDFS(x1,y1,num);}void lemone3(int x,int y,int num)//改变状态开始{int b,c;if(x>=4){return;}for(b=0;b<4;b++){if(c1[b][y]=='-'){c1[b][y]='+';}else{c1[b][y]='-';}}for(b=0;b<4;b++){if(b==y)continue;if(c1[x][b]=='-'){c1[x][b]='+';}else{c1[x][b]='-';}}}void Lemonprintf()//输出步数与路径{int b;printf("%d\n",min);for(b=0;b<min;b++){printf("%d %d\n",y2[b][0]+1,y2[b][1]+1);}}int main(){Lemonscanf();//记录冰箱状态LemonDFS(0,0,0);//开始搜索Lemonprintf();//输出最少步数与路径}