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 | jiangly | 3898 |
2 | tourist | 3840 |
3 | orzdevinwang | 3706 |
4 | ksun48 | 3691 |
5 | jqdai0815 | 3682 |
6 | ecnerwala | 3525 |
7 | gamegame | 3477 |
8 | Benq | 3468 |
9 | Ormlis | 3381 |
10 | maroonrk | 3379 |
# | User | Contrib. |
---|---|---|
1 | cry | 168 |
2 | -is-this-fft- | 165 |
3 | Dominater069 | 161 |
4 | Um_nik | 160 |
5 | atcoder_official | 159 |
6 | djm03178 | 157 |
7 | adamant | 153 |
8 | luogu_official | 150 |
9 | awoo | 149 |
10 | TheScrasse | 146 |
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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