2020-2021:teams:farmer_john:2020牛客暑期多校第十场
| 两侧同时换到之前的修订记录 前一修订版 后一修订版 | 前一修订版 | ||
|
2020-2021:teams:farmer_john:2020牛客暑期多校第十场 [2020/08/14 11:01] 2sozx [总结] |
2020-2021:teams:farmer_john:2020牛客暑期多校第十场 [2020/10/07 21:28] (当前版本) jjleo [比赛名称] |
||
|---|---|---|---|
| 行 1: | 行 1: | ||
| - | ======比赛名称====== | + | ======2020牛客暑期多校第十场====== |
| [[https://ac.nowcoder.com/acm/contest/5675|比赛链接]] | [[https://ac.nowcoder.com/acm/contest/5675|比赛链接]] | ||
| =====A.===== | =====A.===== | ||
| 行 15: | 行 15: | ||
| =====C.===== | =====C.===== | ||
| - | **upsolved by ** | + | **upsolved by JJLeo** |
| ====题意==== | ====题意==== | ||
| ====题解==== | ====题解==== | ||
| + | 大致看没毛病，一细想就不会了，我裂开。 | ||
| =====D.===== | =====D.===== | ||
| **solved by 2sozx JJLeo Bazoka13** | **solved by 2sozx JJLeo Bazoka13** | ||
| 行 35: | 行 36: | ||
| 上述25种随从的对应方式优先级全部排列完毕，按优先级模拟即可。 | 上述25种随从的对应方式优先级全部排列完毕，按优先级模拟即可。 | ||
| =====E.===== | =====E.===== | ||
| - | **upsolved by** | + | **solved by JJLeo** |
| ====题意==== | ====题意==== | ||
| + | 水题。 | ||
| ====题解==== | ====题解==== | ||
| + | 二分即可。 | ||
| =====F.===== | =====F.===== | ||
| **upsolved by ** | **upsolved by ** | ||
| 行 57: | 行 59: | ||
| =====I.===== | =====I.===== | ||
| - | **upsolved by ** | + | **solved by Bazoka13** |
| ====题意==== | ====题意==== | ||
| 行 63: | 行 65: | ||
| =====J.===== | =====J.===== | ||
| - | **upsolved by ** | + | **solved by Bazoka13** |
| ====题意==== | ====题意==== | ||
| + | 给出两棵根给定的树，尽量少地重排编号使得两棵树相同 | ||
| ====题解==== | ====题解==== | ||
| + | 考虑用费用流转移,$dp[i][j]$表示第一颗树里的$i$节点的子树变成第二棵树里$j$节点的子树需要的费用，每次对相同子树大小和儿子数量的节点跑一个最小费用最大流，根据最大流和儿子数量判断可行性，$dp[rt_0][rt_1]$即为答案 | ||
| =====记录===== | =====记录===== | ||
| 0min：开局分题\\ | 0min：开局分题\\ | ||
| 行 80: | 行 83: | ||
| =====总结===== | =====总结===== | ||
| * MJX：要提高打代码速度和准确率，虽然C最后想错了但是也没打出来。 | * MJX：要提高打代码速度和准确率，虽然C最后想错了但是也没打出来。 | ||
| + | * ZYF：比较早的时候就想到了J，但是认为复杂度过高遂自闭两个小时，要敢于尝试，减少这种无意义的自闭时间。 | ||
2020-2021/teams/farmer_john/2020牛客暑期多校第十场.1597374089.txt.gz · 最后更改: 2020/08/14 11:01 由 2sozx
