No.3369 Find MyakuMyaku
タグ : / 解いたユーザー数 16
作問者 :
noya2
Story
Let's find MyakuMyaku! Its face parts have turned into rectangles, though...
Problem Statement
There is a $10^9 \times 10^9$ grid. The top-left cell of this grid is called cell $(0,0)$.
For integers $r,c$ between $0$ and $10^9 - 1$, inclusive, the cell reached by moving $r$ cells down and $c$ cells to the right from cell $(0,0)$ is called cell $(r,c)$.
You are given $N$ rectangles $R_1, R_2, \dots , R_N$. $R_i$ is the rectangle whose top-left corner is cell $(h_{i1}, w_{i1}) $ and whose bottom-right corner is cell $(h_{i2},w_{i2})$.
Here, for $i=1,2,\dots,N$, it is guaranteed that $1 \le h_{i1} \le h_{i2} \le H,1 \le w_{i1} \le w_{i2} \le W$ holds.
Solve the following problem for $k = 1,2,...,N$.
Paint the cells contained in at least one of the rectangles $R_1, R_2, \dots , R_k$ black, and paint the cells contained in none of them white.Determine whether there exist $2$ white cells that are not connected.
Here, $2$ white cells are said to be connected if the $2$ white cells $(i_1, j_1),(i_2,j_2)$ satisfy the following condition.
- There exists a sequence of cells $(x_1, y_1), (x_2, y_2), \dots, (x_L, y_L)$ satisfying the following conditions.
- For all $l = 1, 2, \dots, L$, $(x_l, y_l)$ is painted white.
- For all $l = 1, 2, \dots, L-1$, $|x_l - x_{l+1}| + |y_l - y_{l+1}| = 1$ holds.
- $(x_1, y_1) = (i_1, j_1)$
- $(x_L, y_L) = (i_2, j_2)$
Constraints
- $1\le N, H, W \le 10^3$
- $1 \le h_{i1} \le h_{i2} \le H$
- $1 \le w_{i1} \le w_{i2} \le W$
- All input values are integers.
Input
$N \space H \space W$
$h_{11} \space h_{12} \space w_{11} \space w_{12}$
$h_{21} \space h_{22} \space w_{21} \space w_{22}$
$\vdots$
$h_{N1} \space h_{N2} \space w_{N1} \space w_{N2}$
Output
Output $N$ lines. On the $i$-th line, output the answer for the case $k = i$: Yes if there exist $2$ white cells that are not connected, and No otherwise.
Sample
Sample 1
Input
4 4 4 1 4 1 1 1 1 1 4 1 4 4 4 4 4 1 4
Output
No No No Yes
When $k = 1$, the black cells are the $4$ cells $(1,1),(2,1),(3,1),(4,1)$.
In this case, it can be proved that all white cells are connected to each other, so output No.
When $k = 2$, the black cells are the $7$ cells $(1,1),(1,2),(1,3),(1,4),(2,1),(3,1),(4,1)$.
In this case, it can be proved that all white cells are connected to each other, so output No.
When $k = 3$, the black cells are the $10$ cells $(1,1),(1,2),(1,3),(1,4),(2,1),(2,4),(3,1),(3,4),(4,1),(4,4)$.
In this case, it can be proved that all white cells are connected to each other, so output No.
When $k = 4$, the black cells are the $12$ cells $(1,1),(1,2),(1,3),(1,4),(2,1),(2,4),(3,1),(3,4),(4,1),(4,2),(4,3),(4,4)$.
In this case, it can be proved that cell $(0,0)$ and cell $(2,2)$ are not connected. Therefore, output Yes.
Note that you must output Yes when there exist $2$ white cells that are not connected.
Sample 2
Input
4 3 3 1 1 2 2 2 2 1 1 3 3 2 2 2 2 3 3
Output
No No No Yes
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