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…

题面 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 nea…

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…

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…

传送门 解题思路 树形dp,看到数据范围应该能想到是O(nk)级别的算法,进而就可以设出dp状态,dp[x][j]表示以x为根的子树,距离它为i的点的总和,第一遍dp首先自底向上,dp出每个节点的子树中到他距离为j的,转移方程dp[x][j]=dp[u][j-1] ,第二遍dp自顶向下,dp出每个节点父亲那头的距离为j的,转移方程dp[u][j]+=dp[x][j-1]-dp[u][j-2] 代码 //dp[x][j] 以i为根的子树,距离为j的牛的和 //dp[x][j]+=dp[u][j-1…

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…

我记得我调这道题时中耳炎,发烧,于是在学长的指导下过了也没有发题解 发现我自己的思路蛮鬼畜的 常规操作:\(f[i][j]\) 表示到\(i\)的距离为\(j\)的奶牛有多少只,但注意这只是在第二遍dfs之后 在我的第一遍dfs中(就是下面那个叫build的函数),\(f[i][j]\)的含义是在i这课子树中到\(i\)的距离为\(j\)的奶牛有多少只,所以在第一遍dfs的时候,\(f[i][j]\)的状态只会来自它的儿子们 于是在第一遍dfs就有一个异常简单的方程 \[f[i][j]=\sum…

传送门 dp[i][j][0] 表示点 i 在以 i 为根的子树中范围为 j 的解 dp[i][j][1] 表示点 i 在除去 以 i 为根的子树中范围为 j 的解 状态转移就很好写了 ——代码 #include <cstdio> #include <cstring> #include <iostream> #define N 100001 int n, k, cnt; int f[N], dp[N][21][2], head[N], to[N << 1],…

$k$ 十分小,直接暴力维护 $1$~$k$ 的答案即可. 然后需要用父亲转移到儿子的方式转移一下. Code: #include <bits/stdc++.h> #define M 23 #define N 100005 #define setIO(s) freopen(s".in","r",stdin) using namespace std; int n,K,edges; int f[N][M],hd[N],to[N<<1],nex[N…