2D Conveyor Belt
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Farmer John's milk factory can be described by an \(N\) by \(N\)
(\(1 \le N \le 1000\)) grid of cells that contain conveyor belts. Position (\(a,b\))
describes the cell that is in the \(a\)-th row from the top and \(b\)-th column from
the left. There are \(5\) types of cells:
- "L" — the cell is a leftward facing conveyor belt which moves all items on it 1 cell left every time unit.
- "R" — the cell is a rightward facing conveyor belt which moves all items on it 1 cell right every time unit.
- "U" — the cell is an upward facing conveyor belt which moves all items on it 1 cell up every time unit.
- "D" — the cell is a downward facing conveyor belt which moves all items on it 1 cell down every time unit.
- "?" — Farmer John has not built a conveyor belt at that cell yet.
Note that conveyor belts can also move items outside the grid. A cell \(c\) is
unusable if an item placed at cell \(c\) will never exit the
conveyor belt grid (i.e. it will move around in the grid forever).
Initially, Farmer John has not started building the factory so all cells start
out as "?". For the next \(Q\) (\(1 \le Q \le 2 \cdot 10^5\)) days starting from day
\(1\) and ending at day \(Q\), Farmer John will choose a cell that does not
have a conveyor belt and build a conveyor belt at the cell.
Specifically, during the \(i\)-th day, Farmer John will build a conveyor belt of
type \(t_i\) (\(t_i \in {\text{{L,R,U,D}}}\)) at position (\(r_i,c_i\))
(\(1 \le r_i,c_i \le N\)). It is guaranteed that there is no conveyor belt at
position (\(r_i,c_i\)).
After each day, help Farmer John find the minimum number of unusable cells he
can achieve by optimally building conveyor belts on all remaining
cells without a conveyor belt.
Problem credits: Alex Liang
SCORING
- Inputs 4-5: \(N \le 10\)
- Inputs 6-7: \(N \le 40\)
- Inputs 8-13: No additional constraints
Problem credits: Alex Liang
The first line contains \(N\) and \(Q\).
The \(i\)-th of the next \(Q\) lines contains \(r_i\), \(c_i\), and \(t_i\) in that order.
\(Q\) lines, the \(i\)-th of which describing the minimum number of unusable cells
if Farmer John fills optimally builds conveyor belts on all remaining cells that
do not currently have a conveyor belt.
3 5
1 1 R
3 3 L
3 2 D
1 2 L
2 1 U0
0
0
2
3The conveyor belt after the fifth day is shown below.
RL?
U??
?DL
One optimal way to build conveyor belts on the remaining cells is as follows.
RLR
URR
LDL
In this configuration, the cells at (\(1, 1\)), (\(1, 2\)), and (\(2, 1\)) are
unusable.
3 8
1 1 R
1 2 L
1 3 D
2 3 U
3 3 L
3 2 R
3 1 U
2 1 D0
2
2
4
4
6
6
9The conveyor belt after the eighth day is shown below.
RLD
D?U
URL
No matter what conveyor belt Farmer John can build at the center, all cells
will be unusable.
4 13
2 2 R
2 3 R
2 4 D
3 4 D
4 4 L
4 3 L
4 2 U
3 1 D
4 1 R
2 1 L
1 1 D
1 4 L
1 3 D0
0
0
0
0
0
0
0
11
11
11
11
13riseoj 작성
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2D Conveyor Belt
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2D Conveyor Belt