Problem Statement: this Is there a way to prove that if we are not able to connect the vertices to 1 in the greedy order that has been suggested, then there exists no other answer?
Thanks.
# | User | Rating |
---|---|---|
1 | tourist | 3845 |
2 | jiangly | 3707 |
3 | Benq | 3630 |
4 | orzdevinwang | 3573 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | jqdai0815 | 3532 |
8 | ecnerwala | 3501 |
9 | gyh20 | 3447 |
10 | Rebelz | 3409 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 171 |
2 | adamant | 163 |
3 | awoo | 161 |
4 | nor | 152 |
4 | maroonrk | 152 |
6 | -is-this-fft- | 151 |
7 | TheScrasse | 149 |
8 | atcoder_official | 145 |
8 | Petr | 145 |
10 | pajenegod | 144 |
Name |
---|
Auto comment: topic has been updated by Flvx (previous revision, new revision, compare).
Let's call sum of Ak as Sk
If we can connect (i,j) (i,j != 1), it means Si + Sj >= i * j * c
If Si > Sj, then Si + Si >= i * j * c, Si >= i * (j/2) * c
j/2 >= 1, so Si >= i * 1 * c, We can connect (i, 1) and (1, j).
Aah, got it. Thanks