题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Each cow lives in one of N (1 <= N <= 100,000) different ba…
题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…
题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…
题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…
题目大意:给你一棵树,每个点有点权,边有边权,求一个点,使得其他所有点到这个点的距离和最短,输出这个距离 题解:树形$DP$,思路清晰,转移显然 卡点:无 C++ Code: #include <cstdio> #include <algorithm> #define maxn 100010 const long long inf = 0x3f3f3f3f3f3f3f3f; int head[maxn], cnt; struct Edge { int to, nxt, w; } e…
P2986 [USACO10MAR]伟大的奶牛聚集Great Cow Gat… 题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Each cow lives in on…
题目链接 先把\(1\)作为根求每个子树的\(size\),算出把\(1\)作为集会点的代价,不难发现把集会点移动到\(u\)的儿子\(v\)上后的代价为原代价-\(v\)的\(size\)*边权+(总的\(size\)-\(v\)的\(size\))*边权 #include<iostream> #include<cstring> #include<cstdio> #define int long long using namespace std; const int…
题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…
题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…
题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…
每个点有重数,求到所有点距离最小的点 就是魔改的重心了 #include <bits/stdc++.h> using namespace std; #define int long long const int N = 1000005; vector <pair<int,int> > g[N]; int siz[N],f[N],vis[N],sum[N],c[N],n,m,t1,t2,t3,tot; void dfs1(int p) { vis[p]=1; siz[p]…
有点权的重心,拆掉点dfs不就是了吗 //#include <iostream> #include <cstdio> #include <cstring> //#include <algorithm> //#include <cmath> #define R(a,b,c) for(register int a = (b); a <= (c); ++ a) #define nR(a,b,c) for(register int a = (b);…
题面 题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Each cow lives in one of N (1 <= N <= 100,000) different…
题目描述 Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. Some cows really like their current barn, and some are not so happy. FJ would like to rearrange the cows such that…
传送门 一个深度为i的树可以由一个根节点外加两个深度为i-1的树组成,这就决定了DP该怎么写. 然而我真的没有想到. f[i][j]表示深度为i节点数为j的个数 sum[i][j]表示深度小于等于i节点树为j的个数 #include <cstdio> #define N 402 #define p 9901 int n, m; int f[N][N], sum[N][N]; //f[i][j]表示深度为i节点数为j的个数 //sum[i][j]表示深度<=i节点数为j的树的个数 int…
