P2872 [USACO07DEC]道路建设Building Roads kruskal求最小生成树. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #define re register using namespace std; typedef double ld; t…

https://www.luogu.org/problemnew/show/P2872 算是比较裸的并查集了,已经有路的两个点之间建一条代价为0的边,路径长度计算两点之间的距离,做并查集就好咯. #include <algorithm> #include <iostream> #include <cstring> #include <cstdio> #include <cmath> using namespace std; #define Ldo…

P2872 [USACO07DEC]道路建设Building Roads 题目描述 Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms…

P2872 传送门 首先 题目概括:题目让着求使所有牧场都联通.需要修建多长的路. 显然这是一道最小生成树板子题(推荐初学者做). 那我就说一下kruskal吧. Kruskal算法是一种用来查找最小生成树的算法,由Joseph Kruskal在1956年发表. 用来解决同样问题的还有Prim算法和Boruvka算法等.三种算法都是贪心算法的应用. 和Boruvka算法不同的地方是,Kruskal算法在图中存在相同权值的边时也有效.------- 来自于百度百科 一.基本思路 kruskal利用…

P2872 [USACO07DEC]道路建设Building Roads 题目描述 Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms…

题目描述 Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms. Each of the N (1 ≤ N ≤ 1,000) farms…

\(problem\) 错的原因是\(RE\)(大雾 , 时刻谨记 \(N\) 个地方的话 保守开 \(\frac{N^2}{2}\) 大小. 因为是边. 边最多的情况即完全图 : $1+2+3+4...+n = \frac{N*(N-1)}{2} $ 所以还是个板子. 忽略丑陋的\(2^{18}\) #include <bits/stdc++.h> using namespace std; typedef long long LL; inline LL read () { LL res =…

这道题真的是令人窒息,Kruskal调了贼久一直RE,最后发现数组大小稍微少了那么一点点.(也就10倍吧..) 言归正传,根据本人的分析(以及算法标签的提示),这是一道求最小生成树的题目,当然要注意已经有一些路被建成了,因此他们直接标0即可. 下面是这道题用到了的所有(全局)变量. maxn, n, m就不解释了. x[]和y[]是用来储存农场的坐标的,当然也可以用二维数组写,只是我懒得敲那么多字(说起来差别也不大). f是并查集中储存祖先的数组. 如果有不了解的并查集的可以参考这一片讲解,个人…

题目:洛谷P2872.POJ3625. 题目大意:给你n个点的坐标,有些点已经有边连通,现在要你连上剩下的所有点,求这些边的最小长度是多少(不包括原来的边). 解题思路:最小生成树,把所有边处理出来,跑Kruskal即可.注意原来有的边优先级最高且长度不加进答案.由于边的总数是$n^2$级别的,所以时间复杂度$O(n^2\log n^2)$. C++ Code: #include<cstdio> #include<cstring> #include<algorithm>…

洛谷 P2872 [USACO07DEC]道路建设Building Roads 洛谷传送门 JDOJ 2546: USACO 2007 Dec Silver 2.Building Roads JDOJ传送门 Description Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any o…