P3047 [USACO12FEB]附近的牛Nearby Cows 题目描述 Farmer John has noticed that his cows often move between nearby fields. Taking this into account, he wants to plant enough grass in each of his fields not only for the cows situated initially in that field, but…

题意 给出一棵n个点的无根树,每个点有权值,问每个点向外不重复经过k条边的点权和 题解 设f[i][j]表示所有离i节点距离为j的点权和,v为它周围相邻的点,t为v的个数,则 j > 2 f[i][j] = (sigma f[v][j - 1]) - (t - 1) * f[i][j - 2] j==2 f[i][j] = (sigma f[v][j - 1]) - t * f[i][j - 2] 枚举j,再Dfs即可. 常数巨大的丑陋代码 # include <stdio.h> # i…

题面 比较简单的树形dp(递推?) 设$dp[i][j]$表示距离$i$距离为$j$的点的数目,先预处理$g[i][j]$表示点$i$的子树中距离这个点距离为$j$的点的数目(猫老师讲过,用一个栈维护一下就好了),然后再预处理根节点,之后开始考虑dp. 当进入一个儿子时,首先这个儿子对于所有距离$dis$会继承距离父亲$dis-1$的点,然后是它的子树中距离它为$dis$的点,但是这样距离它为$dis-2$的点会被算重复,减掉就好了 #include<cstdio> #include<c…

P3047 [USACO12FEB]附近的牛Nearby Cows 农民约翰已经注意到他的奶牛经常在附近的田野之间移动.考虑到这一点,他想在每一块土地上种上足够的草,不仅是为了最初在这片土地上的奶牛,而且是为了从附近的田地里去吃草的奶牛. 具体来说,FJ的农场由N块田野构成(1 <= n <= 100,000),每两块田野之间有一条无向边连接(总共n-1条边).FJ设计了农场,任何两个田野i和j之间,有且只有一条路径连接i和j.第 i块田野是C(i)头牛的住所,尽管奶牛们有时会通过k条路到达其…

P3047 [USACO12FEB]附近的牛Nearby Cows 题目描述 Farmer John has noticed that his cows often move between nearby fields. Taking this into account, he wants to plant enough grass in each of his fields not only for the cows situated initially in that field, but…

题意描述 Sightseeing Cows G 给定一张有向图,图中每个点都有点权 \(a_i\),每条边都有边权 \(e_i\). 求图中一个环,使 "环上个点权之和" 除以 "环上各边权之和" 最大.输出最大值. 解释一下,原题目中并没有点明这是一个环,但是从: 奶牛们不会愿意把同一个建筑物参观两遍. 可以看出,不管怎么走,走环一定是最优的,因为重复走相当于无故增加分母. 算法分析 据说这是一道 0/1 分数规划的题目,但是其实可以用更加通俗易懂的方法来解释.…

题目 [USACO14MAR]Counting Friends G 题解 这道题我们可以将 \((n+1)\) 个边依次去掉,然后分别判断去掉后是否能满足.注意到一点, \(n\) 个奶牛的朋友之和必定为偶数,所以去掉的那个数值的奇偶性必定与 \((n+1)\) 个数值之和的奇偶性相同. 接下来很明显的,尽量将朋友多的和朋友多的匹配,所以先从大到小排序,将第一个奶牛和后面的奶牛依次匹配,如果匹配结束,第一个奶牛还有剩余,则此情况必然不可能成立:否则匹配完之后再按照 \(O(n)\) 复杂度的归并…

题目 [USACO14MAR]Sabotage G 题解 本蒟蒻又来了,这道题可以用二分答案来解决.我们可以设答案最小平均产奶量为 \(x \ (x \in[1,10000])\) .然后二分搜索 \(x\) 的最小值. \[\frac{sum-sum[l,r]}{n-(r-l+1)}\leq x \] \[nx-(r-l+1)x\geq sum-sum[l,r] \] \[sum-nx \leq \sum\limits_{i=l}^r{(a[i]-x)} \] 对于如何求 \(\sum\lim…

题目描述 Farmer John has noticed that his cows often move between nearby fields. Taking this into account, he wants to plant enough grass in each of his fields not only for the cows situated initially in that field, but also for cows visiting from nearby…

https://www.luogu.org/problemnew/show/P304 1 #include <bits/stdc++.h> 2 #define up(i,l,r) for(register int i = (l); i <= (r); ++i) 3 #define dn(i,l,r) for(register int i = (l); i >= (r); --i) 4 #define ll long long #define re register using na…