题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1301

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 maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.

The input consists of one to 100 data sets,
followed by a final line containing only 0. Each data set starts with a line
containing only a number n, which is the number of villages, 1 < n < 27,
and the villages are labeled with the first n letters of the alphabet,
capitalized. Each data set is completed with n-1 lines that start with village
labels in alphabetical order. There is no line for the last village. Each line
for a village starts with the village label followed by a number, k, of roads
from this village to villages with labels later in the alphabet. If k is greater
than 0, the line continues with data for each of the k roads. The data for each
road is the village label for the other end of the road followed by the monthly
maintenance cost in aacms for the road. Maintenance costs will be positive
integers less than 100. All data fields in the row are separated by single
blanks. The road network will always allow travel between all the villages. The
network will never have more than 75 roads. No village will have more than 15
roads going to other villages (before or after in the alphabet). In the sample
input below, the first data set goes with the map above.

The output is
one integer per line for each data set: the minimum cost in aacms per month to
maintain a road system that connect all the villages. Caution: A brute force
solution that examines every possible set of roads will not finish within the
one minute time limit.

题意描述:n个城市之间有一些道路连接和维修道路所需的花费,要求n个城市保持直接或间接道路通畅,求出最少花费。

算法分析:赤果果的最小生成树。(最近的切身体会:算法基础一定要扎实,不然高级一点的算法学了之后反倒基础全忘完了)

 #include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std; int g[][],vis[];
int n; int Prim()
{
int t=n-;
int sum=;
while (t--)
{
int k,Min=inf;
for (int i= ;i<n ;i++)
{
if (!vis[i] && Min>g[][i])
{
k=i;
Min=g[][i];
}
}
vis[k]=;
sum += Min;
for (int i= ;i<n ;i++)
{
if (!vis[i] && g[k][i]<g[][i])
g[][i]=g[k][i];
}
}
return sum;
} int main()
{
while (scanf("%d",&n)!=EOF && n)
{
char str[],s[];
int num,cnt;
for (int i= ;i< ;i++)
{
for (int j= ;j< ;j++)
g[i][j]=inf;
}
memset(vis,,sizeof(vis));
for (int i= ;i<n ;i++)
{
scanf("%s%d",str,&num);
int x=str[]-'A';
while (num--)
{
scanf("%s%d",s,&cnt);
int y=s[]-'A';
g[x][y]=g[y][x]=cnt;
}
}
printf("%d\n",Prim());
}
return ;
}

hdu 1301 Jungle Roads 最小生成树的更多相关文章

  1. POJ 1251 && HDU 1301 Jungle Roads (最小生成树)

    Jungle Roads 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/A http://acm.hust.edu.cn/vju ...

  2. Hdu 1301 Jungle Roads (最小生成树)

    地址:http://acm.hdu.edu.cn/showproblem.php?pid=1301 很明显,这是一道“赤裸裸”的最小生成树的问题: 我这里采用了Kruskal算法,当然用Prim算法也 ...

  3. hdu 1301 Jungle Roads krusckal,最小生成树,并查集

    The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was s ...

  4. HDU 1301 Jungle Roads (最小生成树,基础题,模版解释)——同 poj 1251 Jungle Roads

    双向边,基础题,最小生成树   题目 同题目     #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include<stri ...

  5. POJ 1251 + HDU 1301 Jungle Roads 【最小生成树】

    题解 这是一道裸的最小生成树题,拿来练手,题目就不放了 个人理解  Prim有些类似最短路和贪心,不断找距当前点最小距离的点 Kruskal类似于并查集,不断找最小的边,如果不是一棵树的节点就合并为一 ...

  6. 最小生成树 || HDU 1301 Jungle Roads

    裸的最小生成树 输入很蓝瘦 **并查集 int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); } 找到x在并查集里的根结点,如果 ...

  7. POJ 1251 & HDU 1301 Jungle Roads

    题目: Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign ...

  8. hdu 1301 Jungle Roads

    http://acm.hdu.edu.cn/showproblem.php?pid=1301 #include <cstdio> #include <cstring> #inc ...

  9. poj 1251 Jungle Roads (最小生成树)

    poj   1251  Jungle Roads  (最小生成树) Link: http://poj.org/problem?id=1251 Jungle Roads Time Limit: 1000 ...

随机推荐

  1. Elipse安装Spring Tool Suite

    STS实际上是对Eclipse的Spring包装,下载STS IDE可以直接开发spring web. 但大多数人还是喜欢使用eclipse.下面就eclipse安装sts插件做个介绍. 1.首先到s ...

  2. Spark RDD的依赖解读

    在Spark中, RDD是有依赖关系的,这种依赖关系有两种类型 窄依赖(Narrow Dependency) 宽依赖(Wide Dependency) 以下图说明RDD的窄依赖和宽依赖 窄依赖 窄依赖 ...

  3. PHP file_get_contents于curl性能效率比较

    说明大部分内容整理来源于网络,期待你的补充.及不当之处的纠正: 1)fopen/file_get_contents 每次请求远程URL中的数据都会重新做DNS查询,并不对DNS信息进行缓存.但是CUR ...

  4. Android消息推送完美方案[转]

    转自 Android消息推送完美方案 推送功能在手机应用开发中越来越重要,已经成为手机开发的必须.在Android应用开发中,由于众所周知的原因,Android消息推送我们不得不大费周折.本文就是用来 ...

  5. FileOutputSream

    package cd.itcast.fileinputstream; import java.io.File; import java.io.FileNotFoundException; import ...

  6. PHP+MYSQL会员系统的开发实例教程

    本文通过一个简单的实例完成了完整的PHP+MySQL会员系统功能.是非常实用的一个应用.具体实现步骤如下: 一.会员系统的原理: 登陆-->判断-->保持状态(Cookie或Session ...

  7. 集合删数 (vijos 1545) 题解

    [问题描述] 一个集合有如下元素:1是集合元素:若P是集合的元素,则2 * P +1,4*P+5也是集合的元素,取出此集合中最小的K个元素,按从小到大的顺序组合成一个多位数,现要求从中删除M个数位上的 ...

  8. 共享内存shared pool (3):Library cache

    Shared pool物理层面上由许多内存块(chunck)组成.从逻辑功能划分,Shared pool主要由三部分组成:Library cache,Dictionary cache和Control ...

  9. Learning Scrapy笔记(五)- Scrapy登录网站

    摘要:介绍了使用Scrapy登录简单网站的流程,不涉及验证码破解 简单登录 很多时候,你都会发现你需要爬取数据的网站都有一个登录机制,大多数情况下,都要求你输入正确的用户名和密码.现在就模拟这种情况, ...

  10. Python核心编程--学习笔记--3--Python基础

    本章介绍基本的Python语法.编程风格:并简要介绍标识符.变量和关键字,以及变量占用内存的分配和回收:最后给出一个较大的Python样例程序来体验这些特性. 1 语句和语法 1.1 注释 可以在一行 ...