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Way to solve Codeforces Round#416 problem E in O(input) steps
I'm sorry about my poor English.
First, make every cross a node. Give each node a position $$$(x, y)$$$. There is an edge between two nodes if and only if:
- $$$|x_1-x_2|+|y_1-y_2|=1$$$ - The edge splits two different characters.
For example, for the example situation in the problem.
We will get the graph: 
| Rev. | Lang. | By | When | Δ | Comment | |
|---|---|---|---|---|---|---|
| en16 |
|
fast_photon | 2024-06-30 07:03:15 | 4 | ||
| en15 |
|
fast_photon | 2024-06-30 07:02:06 | 48 | ||
| en14 |
|
fast_photon | 2024-06-30 07:00:22 | 20 | #462 -> #416 | |
| en13 |
|
fast_photon | 2024-06-30 06:58:21 | 4 | There's a mistake in spoiler (published) | |
| en12 |
|
fast_photon | 2024-06-30 06:57:26 | 0 | Tiny change: '811E or #462 E in $\ma' -> '811E or #416 E in $\ma' (saved to drafts) | |
| en11 |
|
fast_photon | 2024-06-30 06:56:20 | 0 | (published) | |
| en10 |
|
fast_photon | 2024-06-30 06:50:04 | 5954 | ||
| en9 |
|
fast_photon | 2024-06-30 06:44:28 | 45 | ||
| en8 |
|
fast_photon | 2024-06-30 06:37:26 | 8 | ||
| en7 |
|
fast_photon | 2024-06-30 06:35:40 | 15 | ||
| en6 |
|
fast_photon | 2024-06-30 06:35:15 | 4 | ||
| en5 |
|
fast_photon | 2024-06-30 06:34:44 | 4 | Tiny change: '_1-y_2|=1$\n- The ed' -> '_1-y_2|=1$ \n- The ed' | |
| en4 |
|
fast_photon | 2024-06-30 06:34:04 | 14130 | ||
| en3 |
|
fast_photon | 2024-06-28 15:53:29 | 406 | ||
| en2 |
|
fast_photon | 2024-06-28 14:59:54 | 92 | ||
| en1 |
|
fast_photon | 2024-06-28 14:49:37 | 437 | Initial revision (saved to drafts) |
