Connect the Cities Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3371 Description In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connect…

Connect the Cities Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 701 Accepted Submission(s): 212   Problem Description In 2100, since the sea level rise, most of the cities disappear. Though som…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3371 Problem Description In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The governmen…

Connect the Cities Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 13722    Accepted Submission(s): 3711 Problem Description In 2100, since the sea level rise, most of the cities disappear. Tho…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=3371 题目不难 稍微注意一下 要把已经建好的城市之间的花费定义为0,在用普通Prim算法就可以了:我没有用克鲁斯卡尔算法(Kruskal's algorithm),因为这题数据比较大,而且要处理大量的数据使它为0,怕超时T^T..... #include<iostream> #include<cstdio> #include<algorithm> using namespa…

这个时间短 700多s #include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; struct node{ int u; int v; int w; }que[100000]; int father[505]; bool cmp(struct node a,struct node b){ return a.w<b.w;…

Connect the Cities Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18322 Accepted Submission(s): 4482 Problem DescriptionIn 2100, since the sea level rise, most of the cities disappear. Though some…

Connect the Cities Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9985    Accepted Submission(s): 2843 Problem Description In 2100, since the sea level rise, most of the cities disappear. Thou…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=3371 984ms风险飘过~~~ /************************************************************************/ /* hdu Connect the Cities 最小生成树 题目大意:最小生成树,题目很长,题意很简单就是最小生成树.关键是构建图 */ /****************************************…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3371 Connect the Cities Description In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The…

Connect the Cities Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12903    Accepted Submission(s): 3549 Problem Description In 2100, since the sea level rise, most of the cities disappear. Thou…

Connect the Cities Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 4   Accepted Submission(s) : 3 Problem Description In 2100, since the sea level rise, most of the cities disappear. Though some…

Connect the Cities Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7338    Accepted Submission(s): 2093 Problem Description In 2100, since the sea level rise, most of the cities disappear. Thou…

The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 39561   Accepted: 14444 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undir…

注意输入的数据分别是做什么的就好.还有,以下代码用C++交可以过,而且是500+ms,但是用g++就会TLE,很奇怪. #include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> #include <cmath> #include <cctype> #include <algorithm> #include <numeric&…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4756 题目意思: n-1个宿舍,1个供电站,n个位置每两个位置都有边相连,其中有一条边不能连,求n个位置连通的最小花费的最大值. 解题思路: 和这道题hdu-4126差不多,不过这题不能去掉与供电站相连的边.不同的是这题是一个完全图,求MST时,用kruscal算法的时间复杂度为elge很高会超时,用prim算法复杂度为n^2,所以选用prim算法. PS: double类型的不能用memset,…

本文链接:http://www.cnblogs.com/Ash-ly/p/5409904.html 普瑞姆(Prim)算法: 假设N = (V, {E})是连通网,TE是N上最小生成树边的集合,U是是顶点集V的一个非空子集,算法从U = {uo}(u0 属于 V),TE = {}开始,重复执行下述动作: 在所有u属于U,v属于V - U的边(u, v),且(u, v)属于E中找一条代价最小的边(u0, v0)并并入集合TE中,同时v0并入U,直至U = V为止.此时TE中必有n - 1条边,则T…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=586 题目大意: QS是一种生物,要完成通信,需要设备,每个QS需要的设备的价格不同,并且,这种设备只能在两个QS之间用一次,也就是说,如果一个QS需要和3个QS通信的话,它就必须得买3个设备,同时,对方三个也必须买对应的适合自己的设备.同时,每两个QS之间是有距离的,要完成通信还需要网线,给出每两个QS之间的网线的价值.求一棵生成树,使得所需要的费用最少.数据范围:所有数据都…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3367 题目大意: 让你求最小生成树,并且按照字典序输出哪些点连接.无解输出-1 这里的字典序定义为:(不翻译啦~,详见我的比较函数) A solution A is a line of p integers: a1, a2, ...ap. Another solution B different from A is a line of q integers: b1, b2,…

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=3371 思路: 这道题很明显是一道最小生成树的题目,有点意思的是,它事先已经让几个点联通了.正是因为它先联通了几个点,所以为了判断连通性 很容易想到用并查集+kruskal. 不过要注意 这题有一个坑点,就是边数很多 上限是25000,排序的话可能就超时了.而点数则比较少 上限是500,所以很多人选择用Prim做.但我个人觉得这样连通性不好判断.其实边数多没关系,我们只要去重就好啦,用邻接矩阵存下两点…

Prim 算法属于贪心算法. #include <stdio.h> #define VERTEXNUM 7 #define INF 10000 typedef struct Graph { int vertex[VERTEXNUM]; int edge[VERTEXNUM][VERTEXNUM]; } Graph; void initGraph(Graph* G) { int i, j; int init[][3] = {{1, 2, 10}, {1, 3, 8}, {1, 6, 20}, {…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=3371 其实就是最小生成树,但是这其中有值得注意的地方:就是重边.题目没有告诉你两个城市之间只有一条路可走,所以两个城市之间可能有多条路可以走. 举例: 输入可以包含 1 2 3  // 1到2的成本为3   1 2 5  //1到2的成本为5      因此此时应选成本较低的路. 然后,已经连通的城市之间的连通成本为0. 这题用G++提交得到984ms的反馈,用C++提交则得到484ms的反馈. 很想知…

解题报告:有n个点,然后有m条可以添加的边,然后有一个k输入,表示一开始已经有k个集合的点,每个集合的点表示现在已经是连通的了. 还是用并查集加克鲁斯卡尔.只是在输入已经连通的集合的时候,通过并查集将该集合的点标记到一起,然后剩下的就可以当成是普通的最小生成树来做了题目代码如下: #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cstd…

题目描述: 有n个小岛,其中有的小岛之间没有通路,要修这样一条通路需要花费一定的钱,还有一些小岛之间是有通路的.现在想把所有的岛都连通起来,求最少的花费是多少. 输入: 第一行输入T,代表多少组数据. 第二行输入三个整数n , m , k,分别代表一共有n个小岛,m条路径可供选择,k表示有连通岛的个数. 接下来的m行,每行三个整数p , q , c ,代表建设p到q的通路需要花费的钱为c. 接下的k行,每一行的数据输入如下:先输入Q,代表这个连通分量里面有Q个小岛,然后输入Q个小岛,代表这Q个小…

//我看过Discuss说不能用克鲁斯卡尔因为有很多边 //但是只能用G++过,C++的确超时 #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; struct node { int a, b, cost; }c[]; ]; void init(int n) { ; i <= n; i++) fa[i] = i; } bool cmp(node x, no…

http://acm.hdu.edu.cn/showproblem.php?pid=3371 AC代码: /** /*@author Victor /* C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string…

转自:http://blog.csdn.net/shahdza/article/details/7779230 [HDU]1213 How Many Tables 基础并查集★1272 小希的迷宫 基础并查集★1325&&poj1308 Is It A Tree? 基础并查集★1856 More is better 基础并查集★1102 Constructing Roads 基础最小生成树★1232 畅通工程 基础并查集★2120 Ice_cream's world I 基础并查集★212…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 Kruskal算法 日期 题目地址:https://leetcode-cn.com/problems/connecting-cities-with-minimum-cost/ 题目描述 There are N cities numbered from 1 to N. You are given connections, where each conne…

Problem Description In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these c…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3371 Connect the Cities Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10313    Accepted Submission(s): 2937 Problem Description In 2100, since th…