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…

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…

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…

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…

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…

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…

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…

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…

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: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16070    Accepted Submission(s): 4177 Problem Description In 2100, since the sea level rise, most of the cities disappear. Thou…

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…

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;…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2489 Problem Description For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation. Given a complete graph of n nodes with all nodes and edges…

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…

题目链接: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): 14172    Accepted Submission(s): 5402 Problem Description There are N villa…

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…

HDU.1233 还是畅通工程(Prim) 题意分析 首先给出n,代表村庄的个数 然后出n*(n-1)/2个信息,每个信息包括村庄的起点,终点,距离, 要求求出最小生成树的权值之和. 注意村庄的编号从1开始即可 直接跑prim 代码总览 #include <bits/stdc++.h> #define nmax 105 #define inf 1e8+7 using namespace std; int mp[nmax][nmax]; int n; int totaldis =0; void…

Connect the Cities Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15569    Accepted Submission(s): 4108 Problem Description In 2100, since the sea level rise, most of the cities disappear. Thou…

Agri-Net Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 46319   Accepted: 19052 Description Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He nee…

Battle over Cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 467    Accepted Submission(s): 125 Problem Description It is vitally important to have all the cities connected by highways in…

//归并排序递归方法实现 #include <iostream> #include <cstdio> using namespace std; #define maxn 1000005 int a[maxn], temp[maxn]; long long ans; void MergeSort(int a[], int l, int mid, int r) { ; int i = l, n = mid, j = mid, m = r; while ( i<n &&am…

matrix.c #include <stdio.h> #include <stdlib.h> #include <stdbool.h> #include <limits.h> #include "aqueue.h" #define MAX_VALUE INT_MAX #define MAX_NUM 100 typedef char node_type; typedef struct matrix { node_type vertex[M…

最小生成树,普利姆算法． 简述算法: 先初始化一棵只有一个顶点的树,以这一顶点开始,找到它的最小权值,将这条边上的令一个顶点添加到树中 再从这棵树中的所有顶点中找到一个最小权值(而且权值的另一顶点不属于这棵树) 重复上一步．直到所有顶点并入树中． 图示: 注:以a点开始,最小权值为1,另一顶点是c,将c加入到最小生成树中．树中 a-c 在最小生成树中的顶点找到一个权值最小且另一顶点不在树中的,最小权值是4,另一个顶点是f,将f并入树中, a-c-f 重复上一步骤,a-c-f-d, a-c-f-d…

本文摘自:http://www.cnblogs.com/biyeymyhjob/archive/2012/07/30/2615542.html 最小生成树-Prim算法和Kruskal算法 Prim算法 1.概览 普里姆算法(Prim算法),图论中的一种算法,可在加权连通图里搜索最小生成树.意即由此算法搜索到的边子集所构成的树中,不但包括了连通图里的所有顶点(英语:Vertex (graph theory)),且其所有边的权值之和亦为最小.该算法于1930年由捷克数学家沃伊捷赫·亚尔尼克(英语:…

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…

首先解释什么是最小生成树,最小生成树是指在一张图中找出一棵树,任意两点的距离已经是最短的了. 算法要点: 1.用book数组存放访问过的节点. 2.用dis数组保存对应下标的点到树的最近距离,这里要注意,是对树最近的距离,而不是源点,这和单源最短路径是有区别的. 3.用maps数组保存边的关系. 4.每次先找到离树最近的且没有被访问过的点,以这点的所有边去更新dis数组,也就是更新和树的最近距离. 算法模型: for(循环n-1次,因为默认1点为起始点,已经被访问了) { for(循环n次.)…

最小生成树prim算法实现 所谓生成树,就是n个点之间连成n-1条边的图形.而最小生成树,就是权值(两点间直线的值)之和的最小值. 首先,要用二维数组记录点和权值.如上图所示无向图: int map[7][7];        map[1][2]=map[2][1]=4;        map[1][3]=map[3][1]=2;        ...... 然后再求最小生成树.具体方法是: 1.先选取一个点作起始点,然后选择它邻近的权值最小的点(如果有多个与其相连的相同最小权值的点,随便选取一…