| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 27522 | Accepted: 10625 | Special Judge | ||
Description
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
-+-- ---- ---- -+--
Sample Output
6 1 1 1 3 1 4 4 1 4 3 4 4
题意：4*4的开关图，+表示关，-表示开，只有所有开关状态为-时，灯才会亮，改变开关(i,j)的状态时，i行的开关和j列的开关状态全部改变。
思路：其实和之前棋盘翻转的思路差不多，只不过多加了一个路径记录的问题，自己本来想的是用一个vis数组进行标记，但是，一旦回溯，
所有状态都复原，所以，就用到了一个临时数组每次记录上一步的路径来源，当满足出口条件时，确定最终路径。
只要把《算法入门经典》上p121位向量法给看懂了，这道题的关键部分也解决了，感觉位向量法用在搜索里面特别神奇。
#include<stdio.h>
#include<string.h>
#define N 10
#define inf 0x3f3f3f
int vis[N+N],vis_final[N+N],map[N][N],min;
int Judge()
{
int i,j;
for(i = 0; i < 4; i ++)
for(j = 0; j < 4; j ++)
if(map[i][j] != 1)
return 0;
return 1;
}
void Reverse(int n)
{
int x = n/4,y = n%4,i;
map[x][y] ^= 1;
for(i = 0;i < 4; i ++)
{
if(i != x)
map[i][y] ^= 1;
if(i != y)
map[x][i] ^= 1;
}
return;
}
void dfs(int n,int step)
{
int i;
if(Judge())
{
if(step < min)
{
min = step;
for( i = 1; i <= min; i ++)
{
vis_final[i] = vis[i]; //确定最终路径
}
}
}
if(n > 15)
return ;
dfs(n+1,step);//跳过第n个开关
Reverse(n);//改变第n个灯的状态
vis[step+1] = n;//记录临时路径
dfs(n+1,step+1);//改变第n个开关的状态后，继续往下搜索
Reverse(n);//回溯
}
int main()
{
int i,j;
char str[N];
memset(vis,0,sizeof(vis));
min = inf;
for(i = 0; i < 4; i ++)
{
scanf("%s",str);
for(j = 0; j < 4; j ++)
{
if(str[j] == '+')
map[i][j] = 0;
else
map[i][j] = 1;
}
}
dfs(0,0);
printf("%d
",min);
for(i = 1; i <= min; i ++)
printf("%d %d
",vis_final[i]/4+1,vis_final[i]%4+1);
return 0;
}