2867. Cow Treats

单点时限: 2.0 sec

内存限制: 256 MB



The cows celebrated another banner month for record milk production

and thus have each earned a special treat. They completely fill a

W x H rectangle formation (1 <= W <= 25; 1 <= H <= 25) awaiting

their treat.

Each cow has a unique figure of merit F_rc (1 <= F_rc <= 1,000,000)

which denotes her overall milk production performance. Farmer John

thinks it is only fair to prioritize the treats handed out, rewarding

the highest producing cows first. He plans to traverse the rectangular

formation one row at a time, starting at the beginning of row 1 and

giving out all row 1's treats before he starts on row 2 using the

same method.

He has asked the cows to reorganize themselves so that the cows

with better production are rewarded first. The cows, though, are

not so very good at organization. They can either swap a pair of

rows or swap a pair of columns of their formation. FJ has asked

them to do the best they can by moving the best cow to the upper

left corner (row 1, column 1), the next best cow to row 1, column

2 (if possible), and so on. Of course, the cows can not fully sort

themselves, but they can do their best by following this heuristic:

* determine the order of FJ's treat awards:

1    2   3  ...

W+1 W+2 W+3 ...

* find the highest rated cow; swap rows and columns until she

is at row 1, column 1; never move her again

* Then execute this rule until as many cows are placed as possible:

Find the next highest rated cow. Swap rows and columns

(without moving any higher-rated cow) to move her to the

best possible spot that is still available (e.g., row 1,

column 2 if it's available of row 2, column 1 if no slots

can be achieved in row 1.

By way of example, consider this set of 3 x 4 cows:

5  7  4  1

9 99  2  6

8  3 10 11

The cow with 99 is clearly the highest-rated and belows in the upper

left corner. Swap rows 1 and 2 then swap columns 1 and 2 (or do it

the other way around -- the answer is the same):

99  9  2  6

7  5  4  1

3  8 10 11

The cow with 11 should be rewarded as soon as possible after the

highest-rated cow. She is current in slot (3,4), the last slot to

be rewarded. At this point, it's too late to swap her into the first

row or even the first column. Thus, she needs to move to (2,2) by

swapping columns 2 and 4 then swapping rows 2 and 3:

Swap cols 2 and 4     Swap rows 2 and 3

99   6  2  9          99   6  2  9

7   1  4  5    ->     3  11 10  8

3  11 10  8           7   1  4  5

The cow with 10 is rewarded directly after the cow with 11.  The

cow 9 is already rewarded. The cow with 8 is awarded just after the

cow with 10. The cow with 7 is rewarded directly after the cow with

8.  The cow with 6 is already rewarded. The cow with 5 would best

move to row 3, column 2 but the rows 1 and 2 are frozen as are all

the columns.  Thus, cows 1, 4, and 5 do not move, and the second

diagram above is the "best the cows can do".

Implement this algorithm for other rectangular arrays of cows.

输入格式




* Line 1: Two space-separated integers: W and H

* Lines 2..H+1: Line i+1 contains W space-separated integers F_ic,

where c ranges from 1 to W.

输出格式



* Lines 1..H: Line i contains W space-separated integers representing

the i-th row of cows in the cows' final formation.

样例

Input
4 3
5 7 4 1
9 99 2 6
8 3 10 11
Output
99 6 2 9
3 11 10 8
7 1 4 5

3 人解决,4 人已尝试。

3 份提交通过,共有 27 份提交。

8.3 EMB 奖励。

创建: 8 年,10 月前.

修改: 2 年,4 月前.

最后提交: 6 年,10 月前.

来源: USACO 2011 FEB BRONZE

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