Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensi…

题意: 首先给你一个图,需要你求出最小生成树,首先输入n个节点,用大写字母表示各节点,接着说有几个点和它相连,然后给出节点与节点之间的权值.拿第二个样例举例:比如有3个节点,然后接下来有3-1行表示了边的情况,拿第一行来说:A 2 B 10 C 40表示A有2个邻点,B和C,AB权值是10,AC权值是40. 思路: 直接套prime算法模板和kuskal算法模板,然后处理下数据就可以了. 代码: prime: #include <iostream> #include <cstdio>…

题意: 输入n,然后接下来有n-1行表示边的加边的权值情况.如A 2 B 12 I 25 表示A有两个邻点,B和I,A-B权值是12,A-I权值是25.求连接这棵树的最小权值. 思路: 一开始是在做莫队然后发现没学过最小生成树,就跑过来做模板题了... Kruskal的使用过程:先按权值大小排序,然后用并查集判断是否能加这条边 Kruskal详解博客:[贪心法求解最小生成树之Kruskal算法详细分析]---Greedy Algorithm for MST 考试周还在敲代码...我... upd…

Jungle Roads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19536   Accepted: 8970 Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some…

今天刷一下水题练手入门,明天继续. poj1861 Network(最小生成树)新手入门题. 题意:输出连接方案中最长的单根网线长度(必须使这个值是所有方案中最小的),然后输出方案. 题解:本题没有直接求生成树,但如果连接n个集线器的方案多于n-1条边,那么必存在回路,因此去掉某些边剩下的边和所有顶点构成一个生成树.对于一个图的最小生成树来说,它的最大边满足所有生成树的最大边里最小,正和题意. 吐槽:题目样例是错的... #include<cstdio> #include<cstring…

题目链接: https://vjudge.net/problem/POJ-1251 题目大意: 首先给你一个图,需要你求出最小生成树,输入N个节点,用大写字母表示了节点,然后节点与节点之间有权值. 思路:这里需要编号,其他的就是模板 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue&…

网址:https://vjudge.net/contest/66965#overview 第一题: poj1251 裸最小生成树 #include<iostream> #include<algorithm> #include<cstring> using namespace std; const int maxn=100500; struct node { int x; int y; int w; }a[maxn]; int n,k; char s,v; int cnt…

https://vjudge.net/problem/POJ-1251 The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large roa…

题目链接 http://poj.org/problem?id=1251 题意 有n个村庄,村庄之间有道路连接,求一条最短的路径能够连接起所有村庄,输出这条最短路径的长度. 思路 最小生成树问题,使用普利姆算法(Prime)或者克鲁斯卡尔算法(Kruskal)解决. 代码 Prime算法: #include <algorithm> #include <iostream> #include <cstring> #include <cstdio> using na…

以此图为例: package com.datastruct; import java.util.Scanner; public class TestKruskal { private static class Edge{ public Edge(int begin,int end,int weight){ this.begin = begin; this.end = end; this.weight = weight; } int begin; int end; int weight; publ…