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POJ 2965 枚举+位运算

2015-10-21 17:07 330 查看
The Pilots Brothers’ refrigerator

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 21306 Accepted: 8223 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

由于每个开关只有开和关的两个状态。所以当对一个开关操作奇数次时只相当于操作了1一次,操作了偶数次相当于没有操作。因此每次操作只有不操作和操作0次的区别,那就变成了从16个开关中找几个开关操作的问题。可知i从1遍历到1<<16 -1,每个i中1出现的位置就为对该处开关进行操作。最后判断一下操作后的tem是否为0就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int main(){
char c;
int a = 0;
int i,j,k;
int G[4][4] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
for(i = 0;i<16;i++){
c = getchar();
if(c == '+') a^=1<<i;
if((i+1)%4 == 0) getchar();
}
int lim = 1<<16;
for(i = 0;i<lim;i++){
int tem = a;
for(j = 0;j<16;j++){
if( i&(1<<j) ){
int x = j/4;
int y = j%4;
for(k = 0;k<4;k++){
tem ^= 1<<G[x][k];
tem ^= 1<<G[k][y];
}
tem^=1<<G[x][y];
}
}
if(!tem) break;
}
int sum = 0;

for(j = 0;j<16;j++) if( i&(1<<j) ) sum++;

printf("%d\n",sum);
for(j = 0;j<16;j++)
if( i&(1<<j) ){
int x = j/4;
int y = j%4;
printf("%d %d\n",x+1,y+1);
}

}
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