The Pilots Brothers' refrigerator
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 12909 | Accepted: 4833 | 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
Source
Northeastern Europe 2004, Western Subregion
题解:
高斯消元入门请看POJ1222 点击打开链接POJ1222
这题与之前的题目有所不同 不仅仅求最小步数 而且要输出先后顺序
实际上还是有无穷解要DFS的 所以当DFS出最少解的时候
数组X[i]里记录的值就是方程组的解
通过之前将二维坐标转换为方程组的系数阵的方法 现在知道是第几个的解
在将这个值转化为坐标即可
#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=230;
int a[maxn][maxn+1],x[maxn];//a 是系数矩阵和增广矩阵，x 是最后存放的解
// a[][maxn]中存放的是方程右面的值（bi）
int equ,var;//equ 是系数阵的行数，var 是系数矩阵的列数（变量的个数）
int free_num,ans=100000000;
int abs1(int num) //取绝对值
{
if (num>=0) return num;
else
return -1*num;
}
void Debug(void)
{
int i, j;
for (i = 0; i < equ; i++)
{
for (j = 0; j < var + 1; j++)
{
cout << a[i][j] << " ";
}
cout << endl;
}
cout << endl;
}
//调试输出，看消元后的矩阵值，提交时，就不用了
inline int gcd(int a, int b) //最大公约数
{
int t;
while (b != 0)
{
t = b;
b = a % b;
a = t;
}
return a;
}
inline int lcm(int a, int b) //最小公倍数
{
return a * b / gcd(a, b);
}
int dfs(int p) //枚举自由解，只能取0-1，枚举完就回带，找到最小的
{
if (p<=free_num-1) //深入到了主对角线元素非0 的行了
{
//下面就是回带的代码啊
for(int i = free_num-1; i >= 0; --i)
{
int tmp = a[i][var] % 2;
for(int j = i+1; j < var; ++j) //x[i]取决于x[i+1]--x[var]啊，所以后面的解对前面的解有影
//响。
if(a[i][j] != 0)
tmp = ( tmp - (a[i][j]*x[j])%2 + 2 ) % 2;
x[i] = (tmp/a[i][i]) % 2; //上面的正常解
} //回带完成了
//计算解元素为1 的个数；
int sum=0;
for(int i=0; i<var; i++) sum+=x[i];
if (ans>sum) ans=sum;
return 0;
}
x[p]=0;
dfs(p-1);
x[p]=1;
dfs(p-1);
}
void swap(int &a,int &b)
{
int temp=a; //交换 2 个数
a=b;
b=temp;
}
int Gauss()
{
int k,col = 0;
//当前处理的列
for(k = 0; k < equ && col < var; ++k,++col)
{
int max_r = k;
for(int i = k+1; i < equ; ++i)
if(a[i][col] > a[max_r][col])
max_r = i;
if(max_r != k)
{
for(int i = k; i < var + 1; ++i)
swap(a[k][i],a[max_r][i]);
}
if(a[k][col] == 0)
{
k--;
continue;
}
for(int i = k+1; i < equ; ++i)
{
if(a[i][col] != 0)
{
int LCM = lcm(a[i][col],a[k][col]);
int ta = LCM/a[i][col], tb = LCM/a[k][col];
if(a[i][col]*a[k][col] < 0)
tb = -tb;
for(int j = col; j < var + 1; ++j)
a[i][j] = ( (a[i][j]*ta)%2 - (a[k][j]*tb)%2 + 2 ) % 2;
// 0 和 1 两种状态
}
}
}
//a[i][j]只有
//上述代码是消元的过程，行消元完成
//解下来 2 行，判断是否无解
//注意 K 的值，k 代表系数矩阵值都为 0 的那些行的第 1 行
for(int i = k; i < equ; ++i)
if(a[i][col] != 0)
return -1;
//Debug();
//唯一解或者无穷解,k<=var
//var-k==0 唯一解；var-k>0 无穷多解，自由解的个数=var-k
//能执行到这，说明肯定有解了，无非是 1 个和无穷解的问题。
//下面这几行很重要，保证秩内每行主元非 0，且按对角线顺序排列，就是检查列
for(int i = 0; i <equ; ++i)//每一行主元素化为非零
if(!a[i][i])
{
int j;
for(j = i+1; j<var; ++j)
if(a[i][j])
break;
if(j == var)
break;
for(int k = 0; k < equ; ++k)
swap(a[k][i],a[k][j]);
}
// ----处理保证对角线主元非 0 且顺序，检查列完成
free_num=k;
if (var-k>0)
{
dfs(var-1);
return ans;
//无穷多解，先枚举解，然后用下面的回带代码进行回带；
//这里省略了下面的回带的代码；不管唯一解和无穷解都可以回带，只不过无穷解
//回带时，默认为最后几个自由变元=0 而已。
}
if(var-k<0)
return -1;
// 无解返回 -1
if (var-k==0)//唯一解时
{
//下面是回带求解代码，当无穷多解时，最后几行为 0 的解默认为 0；
for(int i = k-1; i >= 0; --i) //从消完元矩阵的主对角线非 0 的最后 1 行，开始往
//回带
{
int tmp = a[i][var] % 2;
for(int j = i+1; j < var; ++j) //x[i]取决于 x[i+1]--x[var]啊，所以后面的解对前面的解有影响。
if(a[i][j] != 0)
tmp = ( tmp - (a[i][j]*x[j])%2 + 2 ) % 2;
//if (a[i][i]==0) x[i]=tmp;//最后的空行时，即无穷解得
//else
x[i] = (tmp/a[i][i]) % 2; //上面的正常解
}
int sum=0;
for(int i=0; i<var; i++)
sum+=x[i];
return sum;
//回带结束了
}
}
int main()
{
char map[6];
equ=var=16;
while(cin>>map)
{
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
for(int i=0; i<4; i++)
{
if(i)
cin>>map;
for(int j=0; j<4; j++)
{
int k=i*4+j;
a[k][k]=1;
for(int a1=0; a1<4; a1++)
for(int b1=0; b1<4; b1++)
if(a1==i||b1==j)
a[k][a1*4+b1]=1;
if(map[j]=='+')
a[k][16]=1;
}
}
int j1=Gauss();
cout<<j1<<endl;
for(int i=0; i<16; i++)
if(x[i])
cout<<i/4+1<<' '<<i%4+1<<endl;
}
return 0;
}