Question : RRFRNDS I used brute force to solve this and was expecting a TLE but got a WA instead Somebody provide me some hints to solve this
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
4 | atcoder_official | 161 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
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You can find editorial at codechef. Editorial link has to be under the statement as i remember.
Let's call list[i] is all friend of user[i], after that, we check all pair (i, j), if currently i and j is not friend and exists an user in list[i] is also friend of j, then pair (i, j) is valid, we increase the answer to 1.
That's an O(N^3) implementation and will time out