Problem:Minesweeper Master

Google code jam Qualification Round 2014

 题目链接:https://code.google.com/codejam/contest/dashboard?c=2974486#s=p2

Download：source code

word版文档下载

 

#Define LEFT = R*C – M

1. M==R*C-1

ok.

C	*	*
*	*	*
*	*	*

1. M<R*C-1

   1. R==1 or C == 1

ok.

LEFT = {1}

C

*

*

 

C

*

*

1. R==2 or C == 2

It is ok when the LEFT are even, and so does C==2.

LEFT = {1,4,6,8,..2n} n=max(R,C);

 

C	.	*
.	.	*

 

C	.	*
.	*(wrong)	*

1. R>=3 && C >= 3

LEFT = {1,2,3,4,5,6,7,8,9,10,…,i,…,R*C};

LEFT = {1,4, 6,8,9,10,…,i,…,R*C}; 8<=i<=R*C; //Proved at 2.3.1

1

4

C	.	*	*
.	.	*	*
*	*	*	*

6

C	.	.	*
.	.	.	*
*	*	*	*

8

C	.	.	.
.	.	.	.
*	*	*	*

9

C	.	.	*
.	.	.	*
.	.	.	*

10

C	.	.	.
.	.	.	.
.	.	*	*

11

C	.	.	.
.	.	.	.
.	.	.	*

12

C	.	.	.
.	.	.	.
.	.	.	.

1. Prove that i is from 8 to R*C one by one

Each i from 8 to R*C can be

1. firstly

i = 8

C	1	1	*	*	*	*
1	1	1	*	*	*	*
1	1	*	*	*	*	*
*	*	*	*	*	*	*

1. secondly

if we want to goto i(8 < i <= R*C)

1. 8 < i <= 8 + 2 * (R-2)

   1. 0 == i % 2

Only set the first and second columns is enough.

Add two each time.

C	0	1	*	*	*	*
0	0	1	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*
1	1	*	*	*	*	*
*	*	*	*	*	*	*

1. 0 != i % 2

Set the first and second columns to (i-1).

And set [2][2]

C	0	1	*	*	*	*
0	0	1	*	*	*	*
0	1	1	*	*	*	*
0	1	*	*	*	*	*
1	1	*	*	*	*	*
*	*	*	*	*	*	*

1. 8 + 2 * (R-2) < i <= 2 * (R+C-2)

   1. Set full of the first and second columns

C	0	1	*	*	*	*
0	1	1	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*

1. 0 == i % 2

Only set the first and second rows is enough.

Add two each time.

C	0	0	0	1	*	*
0	1	1	1	1	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*

 

1. 0 != i % 2

Set the first and second columns to (i-1).

And set [2][2]

C	0	0	0	1	*	*
0	0	1	1	1	*	*
0	1	1	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*

1. 2 * (R+C-2) < i <= R*C

C	0	0	0	0	0	0
0	1	1	1	1	1	1
0	1	*	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*

1. 2 * (R+C-2) < i <= 2*(R+C-2)+(R-2)*(column-2)

3 <= column <= C;

Each column from the third to the last left (R-2) positions.

1. 2 * (R+C-2) < i <= 2*(R+C-2)+(R-2)*(3-2)

Set the third column (i-2*(R+C-2)) from [2][2] to [R-1][2]

		3
C	0	0	0	0	0	0
0	0	1	1	1	1	1
0	0	1	*	*	*	*
0	1	1	*	*	*	*
0	1	*	*	*	*	*
0	1	*	*	*	*	*

1. 2 * (R+C-2) < i <= 2*(R+C-2)+(R-2)*(4-2)

Set full of the third column.

Set the fourth column (i-2*(R+C-2) – (R-2)) from [2][3] to [R-1][3]

		3	4
C	0	0	0	0	0	0
0	0	0	1	1	1	1
0	0	0	1	*	*	*
0	0	0	1	*	*	*
0	0	1	1	*	*	*
0	0	1	*	*	*	*

1. 2 * (R+C-2) < i <= 2*(R+C-2)+(R-2)*(k-2)

Set full of the third to (k-1) columns.

Set the k column (i-2*(R+C-2) – (k-3)(R-2)) from [2][k-1] to [R-1][k-1]

		3		k-1	k
C	0	0	0	0	0	0
0	0	0	0	0	1	1
0	0	0	0	0	1	*
0	0	0	0	1	1	*
0	0	0	0	1		*
0	0	0	0	1		*