[google大赛 赛前模拟题] HouseParty

摘要:用窗、门和预制的墙板砌尽可能大的屋子。

Problem Statement
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We are building a rectangular house using manufactured sections for the walls. Each section is 4 feet long. There are three types of sections: window sections, door sections, and regular sections. We have a fixed assortment of these sections, and want to design the house to have a maximum area, subject to the following rules:
1) A house side must contain no more than one door.
2) The house must have at least one door.
3) A door section may not be at the end of a wall.
4) Each window section must be adjacent to two regular sections, one on each side of it in its wall.
Create a class HouseParty that contains a method maxArea that is given numReg, numWin, and numDoor (the available quantity of each type of section) and that returns the maximum area of a house built from those sections. You are not required to use all the sections. If no house can be built your method should return 0.
Definition
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Class:
HouseParty
Method:
maxArea
Parameters:
int, int, int
Returns:
int
Method signature:
int maxArea(int numReg, int numWin, int numDoor)
(be sure your method is public)
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Constraints
-
numReg, numWin, and numDoor will each be between 0 and 50 inclusive.
Examples
0)

????
8
0
0
Returns: 0
No house can be built since you have no door sections.
1)

????
8
0
1
Returns: 48
One way is to use 3 regular sections for the north wall, one regular section for the east wall and one for the west wall, and to use a door section between 2 regular sections for the south wall. This gives a house that is 12′ x 4′. Below is a picture (with door and window sections shown as D and W, and regular sections shown as – or | )
 
    —
   |   |
    -D-
2)

????
9
8
2
Returns: 144
One design is:
    -D-
   |   |
   D   W
   |   |
    -W-
3)

????
6
23
13
Returns: 48
We are very short of regular sections; one design is:
    -
   | |
   D W
   | |
    -
 
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