Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19884   Accepted: 8315 Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each…

题意:给出n个村庄之间的距离,再给出已经连通起来了的村庄.求把所有的村庄都连通要修路的长度的最小值. 思路:Kruskal算法 课本代码: //Kruskal算法 #include<iostream> using namespace std; int fa[120]; int get_father(int x){ return fa[x]=fa[x]==x?x:get_father(fa[x]);//判断两个节点是否属于一颗子树(并查集) } int main(){ int n; int p[…

Constructing Roads Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2421 Appoint description:  System Crawler  (2015-05-27) Description There are N villages, which are numbered from 1 to N, and y…

题目链接:http://poj.org/problem?id=2421 实际上又是考最小生成树的内容,也是用到kruskal算法.但稍稍有点不同的是,给出一些已连接的边,要在这些边存在的情况下,拓展出最小生成树来. 一般来说,过到这四组数据大体上就能AC了.  1.题目给出的案例数据    2.连接的道路可能把所有的村庄都已经连通了       3.两个村庄给出多次,即连接这两个村庄的道路是重复的!. 2.3这两种情况的数据如下(为了好看,自己出了一组,当然Sample Input 那组也行):…

Constructing Roads 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/D Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two villa…

Constructing Roads Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road betw…

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a…

题意:要在n个城市之间建造公路,使城市之间能互相联通,告诉每个城市之间建公路的费用,和已经建好的公路,求最小费用. 解法:最小生成树.先把已经建好的边加进去再跑kruskal或者prim什么的. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<math.h> #includ…

给一个n个点的完全图 再给你m条道路已经修好 问你还需要修多长的路才能让所有村子互通 将给的m个点的路重新加权值为零的边到边集里 然后求最小生成树 #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string> #define cl(a,b) memset(a,b,sizeof(a))…

Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or the…

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a…

题意:有几个村庄,要修最短的路,使得这几个村庄连通.但是现在已经有了几条路,求在已有路径上还要修至少多长的路. 分析:用Prim求最小生成树,将已有路径的长度置为0,由于0是最小的长度,所以一定会被Prim选中加入最小生成树. package Map; import java.util.Scanner; /** * Prime */ public class Poj_2421_Prim { static int MAXVEX = 200; static int n, m; static int[…

Constructing Roads There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B,…

http://poj.org/problem?id=2421 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24132   Accepted: 10368 Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can con…

Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: Accepted: Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can…

求最小生成树.有一点点的变化,就是有的边已经给出来了.所以,最小生成树里面必须有这些边,kruskal和prim算法都能够,prim更简单一些.有一点须要注意,用克鲁斯卡尔算法的时候须要将已经存在的边预处理一下,并查集转化为同一个祖先.记得要找他们的祖先再转化.普里姆算法仅仅须要将那些已经存在的边都初始化为0就能够了. kruskal: #include<iostream> #include<cstdlib> #include<cstring> #include<…

时限: 1000MS   内存限制: 10000K 提交总数: 37001   接受: 17398 描述 热带岛屿拉格里山的首长有个问题.几年前,大量的外援花在了村庄之间的额外道路上.但是丛林不断地超越道路,因此庞大的道路网太昂贵而无法维护.老年人理事会必须选择停止维护一些道路.左上方的地图显示了目前正在使用的所有道路,以及每月维护这些道路的费用.当然,即使路线不像以前那么短,也需要采取某种方式在所有村庄之间保持通行.长老院长想告诉长老委员会每月要花多少钱才能维持连接所有村庄的道路.在上面的地图…

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…

题目: 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 expensive to ma…

思路:首先使用二维数组dis[][]处理输入, 对于已经修好的路,将其对应的dis[i][j]置为零即可.最后再将    所有的dis[][]保存到边结构体中,使用Kruskal算法求得最小生成树. #include<iostream> #include<vector> #include<string> #include<cmath> #include<set> #include<algorithm> #include<cstd…

题目链接:problemCode=1372">ZOJ1372 POJ 1287 Networking 网络设计 Networking Time Limit: 2 Seconds      Memory Limit: 65536 KB You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, a…

注意: 注意数组越界问题(提交出现runtimeError代表数组越界) 刚开始提交的时候,边集中边的数目和点集中点的数目用的同一个宏定义,但是宏定义是按照点的最大数定义的,所以提交的时候出现了数组越界问题,以后需要注意啦. Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads betwe…

题目连接:problemId=542" target="_blank">ZOJ 1542 POJ 1861 Network 网络 Network Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Andrew is working as system administrator and is planning to establish a new network in his com…

主题链接:http://poj.org/problem?id=1789 思维:一个一个点,每两行之间不懂得字符个数就看做是权值.然后用kruskal算法计算出最小生成树 我写了两个代码一个是用优先队列写的.可是超时啦,不知道为什么.希望有人能够解答.后面用的数组sort排序然后才AC. code: 数组sort排序AC代码: #include<cstdio> #include<queue> #include<algorithm> #include<iostream…

题目链接:http://poj.org/problem?id=1797 开始题意理解错.不说题意了. 并不想做这个题,主要是想测试kruskal 模板和花式并查集的正确性. 已AC: /* 最小生成树 kruskal算法 过程:每次选取没有参与构造最小生成树并且加入之后不会构成回路的边中权值最小的一条 作为最小生成树的一条新边.直至选择了V-1条边. */ #include <stdio.h> #include <string.h> #include <iostream>…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1102 Constructing Roads Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21947    Accepted Submission(s): 8448 Problem Description There are N villa…

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 14021   Accepted: 5484   Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the c…

/*kruskal算法*/ #include <iostream> //#include <fstream> #include <algorithm> using namespace std; /*708K 922MS*/ typedef struct _edge { int x,y; int w; }edge; int n; int num; //fstream fin; void kruskal(edge *e,int len); int cmp(const voi…

The Unique MST 时间限制: 10 Sec  内存限制: 128 MB提交: 25  解决: 10[提交][状态][讨论版] 题目描述 Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tre…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1102 题意 有n个村庄(编号1~n),给出n个村庄之间的距离,开始时n个村庄之间已经有了q条路,现在需要修一条路,这条路连接起所有的村庄,求在已经存在的路径的基础上,最少还需要修多长的路. 思路 普通最小生成树是从零开始构造一棵最小生成树,而这题开始时图中已经有了一些路径,那这些路就不需要被修了,所以将这些路的修理长度置为0,然后使用Prime算法或者Kruskal算法求解即可. 代码 Prime算…