Poj-2965-the pilots brothers& #39; Refrigerator (High efficiency greedy!!) ）

The pilots ' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19356		Accepted: 7412		Special Judge

Description

The game "The pilots brothers:following the stripy elephant" has a quest where a player needs to open a refrigerator.

There is handles on the refrigerator door. Every handle can is in one of the States:open or closed. The refrigerator is open if all handles is open. The handles is 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 are 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 is several solutions, you could give any one of the them.

Sample Input

-+-----------+--

Sample Output

61 11 31 44 14 34 4

Source

Northeastern Europe 2004, Western subregion

Look at what the ACM practice suggests, and what it says is an enumeration.

But I enumerated the half-day. I didn't get it.

Online search of other people's puzzle, are said to have efficient greedy algorithm. and figured it out for half a day.

Example:

-+--

----

----

----

To change this to a minus, you have to change the number of symbols in the row and column of the + sign all at once (the position of the + sign changes once)

So there are, for example, the following number of changes:

4744

2422

2422

2422

7 is the number of times the + position needs to be changed, and 4 is the number of changes required by the + number in the row and column (not including the + number)

The result is an efficient solution that calculates the number of times the row and column need to be changed when each input encounters ' + '.

When the input is finished, the array is traversed, all the odd positions are the position of the operation, and the sum of the odd positions is the last number of operations.

But such algorithms apply only if n is even.

PS: This problem will not have "impossible" situation.

AC Code:

Poj-2965-the pilots brothers& #39; refrigerator (efficient greedy!!) )