CodeChef April Challenge
qwer
posted @ 2014年5月10日 16:44
in CodeChef
, 1509 阅读
armer Feb |
POTATOES | 5077 | 46.22 |
|
Shortest Path in Binary Trees |
BINTREE | 3863 | 37.89 |
|
Chef and Digits |
ADIGIT | 2678 | 14.21 |
|
Counting Matrices |
CNPIIM | 1927 | 18.16 |
|
Divide the Tangerine |
TANGDIV | 1486 | 17.4 |
|
Sereja and Permutation |
SEAPERM | 1174 | 8.94 |
|
Cards, bags and coins |
ANUCBC | 343 | 5.39 |
|
Final Battle of Chef |
FBCHEF | 57 | 6.87 |
|
Chef and Tree Game |
GERALD08 | 28 | 1.47 |
| Make It Zero 3 | LMATRIX3 | 4 | 1 |
GERALD08题解 -> http://blog.sina.com.cn/s/blog_5a4635700101i70w.html
FBCHER题解 :
首先考虑询问，要求当前时间点到根路径上破产的点的个数，那么只要知道每一个点是在什么时候挂掉的就可以了。
考虑操作，要将离T点距离为d的点的权值减 X/(2^d)。先得到树的bfs序，对于T的子树，同一层的点减少的权值是相同的，从T点向上走，同一层且没有减过的点减少的权值是相同的。维护以bfs序建的线段树即可
再考虑询问，按时间做，如果一个点在这个时刻挂了，就把它的子树减1，这个只要搞个dfs序就可以了
2014年5月11日 08:03
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