B - The Pilots Brothers’ refrigerator
文章目录
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
}