Jungle Roads
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 25993   Accepted: 12181

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 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. 


Input

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. 

Output

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. 

Sample Input

9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0

Sample Output

216
30

Source

——————————————————————————————————————

题目的意思是在给出的一张图中找出一棵最小生成树。

建图后kruskal即可

#include <iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
using namespace std;
#define LL long long struct node
{
int u,v,w;
} p[100005];
int n,cnt,pre[100]; bool cmp(node a,node b)
{
return a.w<b.w;
}
void init()
{
for(int i=0;i<100;i++)
pre[i]=i;
} int fin(int x)
{
return pre[x]==x?x:pre[x]=fin(pre[x]);
} void kruskal()
{
sort(p,p+cnt,cmp);
init();
int cost=0,ans=0,flag=0;
for(int i=0;i<cnt;i++)
{
int a=fin(p[i].u);
int b=fin(p[i].v);
if(a!=b)
{
pre[a]=b;
cost+=p[i].w;
ans++;
}
if(ans==n-1)
{
break;
}
} printf("%d\n",cost);
} int main()
{
char ch1,ch2;
int k,val;
while(~scanf("%d",&n)&&n)
{
cnt=0;
for(int i=0; i<n-1; i++)
{
cin>>ch1>>k;
for(int j=0; j<k; j++)
{
cin>>ch2>>val;
p[cnt].u=ch1-'A',p[cnt].v=ch2-'A',p[cnt++].w=val;
}
}
kruskal();
}
return 0;
}

HDU1301&&POJ1251 Jungle Roads 2017-04-12 23:27 40人阅读 评论(0) 收藏的更多相关文章

  1. Codeforces807 C. Success Rate 2017-05-08 23:27 91人阅读 评论(0) 收藏

    C. Success Rate time limit per test 2 seconds memory limit per test 256 megabytes input standard inp ...

  2. 团体程序设计天梯赛L1-027 出租 2017-03-23 23:16 40人阅读 评论(0) 收藏

    L1-027. 出租 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 陈越 下面是新浪微博上曾经很火的一张图: 一时间网上一片求救声, ...

  3. 利用autotools工具制作从源代码安装的软件 分类: linux 2014-06-02 23:27 340人阅读 评论(0) 收藏

    编写程序(helloworld.c)并将其放到一个单独目录. helloworld.c: #include<stdio.h> int main() { printf("hello ...

  4. 随机L系统分形树 分类: 计算机图形学 2014-06-01 23:27 376人阅读 评论(0) 收藏

    下面代码需要插入到MFC项目中运行,实现了计算机图形学中的L系统分形树. class Node { public: int x,y; double direction; Node(){} }; CSt ...

  5. HDU2033 人见人爱A+B 分类: ACM 2015-06-21 23:05 13人阅读 评论(0) 收藏

    人见人爱A+B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Su ...

  6. 使用URLConnection获取网页信息的基本流程 分类: H1_ANDROID 2013-10-12 23:51 3646人阅读 评论(0) 收藏

    参考自core java v2, chapter3 Networking. 注:URLConnection的子类HttpURLConnection被广泛用于Android网络客户端编程,它与apach ...

  7. pascal矩阵 分类: 数学 2015-07-31 23:01 3人阅读 评论(0) 收藏

    帕斯卡矩阵 1.定义       帕斯卡矩阵:由杨辉三角形表组成的矩阵称为帕斯卡(Pascal)矩阵. 杨辉三角形表是二次项 (x+y)^n 展开后的系数随自然数 n 的增大组成的一个三角形表. 如4 ...

  8. NYOJ-235 zb的生日 AC 分类: NYOJ 2013-12-30 23:10 183人阅读 评论(0) 收藏

    DFS算法: #include<stdio.h> #include<math.h> void find(int k,int w); int num[23]={0}; int m ...

  9. HDU 2035 人见人爱A^B 分类: ACM 2015-06-22 23:54 9人阅读 评论(0) 收藏

    人见人爱A^B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Su ...

随机推荐

  1. 中兴 F412 超级帐号telecomadmin破解(适用2015版h啊RowCount="0") TEWA-300AI EPON TEWA-500AI EPON破解

    1.telnet 192.168.1.1 root/Zte521    有些密码也是root 2.输入sendcmd 1 DB p UserInfo 老本大多数教程会返回超级管理员帐号密码: < ...

  2. Linux操作系统-基本命令(一)

    熟悉Linux命令基础 Linux系统的终端窗口 字符终端为用户提供了一个标准的命令行接口,在字符终端窗口中,会显示一个Shell提示符,通常为$. 用户可以在提示符后输入带有选项和参数的字符命令,并 ...

  3. lambda架构简介

    1.Lambda架构背景介绍 Lambda架构是由Storm的作者Nathan Marz提出的一个实时大数据处理框架.Marz在Twitter工作期间开发了著名的实时大数据处理框架Storm,Lamb ...

  4. 源文件封装为IP的步骤

    因为模块的交接,最好将写好的源文件和生成的IP封装一个IP,然后再转交给其他的同事使用,这是一种好的习惯.但是对于,封装的过程还是需要注意一下.实际的看看步骤吧.1)将源文件和使用到的IP生成工程. ...

  5. 接口自动化(三)--读取json文件中的数据

    上篇讲到实际的请求数据放置在json文件内,这一部分记述一下python读取json文件的实现. 代码如下(代码做了简化,根据需要调优:可做一些容错处理): 1 import json 2 3 cla ...

  6. ubuntu,debian root密码忘记破解

    开机启动的时候在grub引导时,按住e进行启动项编辑,修改开头有linux字符及最后又ro字符的行,将ro字符改为rw single init=/bin/bash按F10键进行启动即可进入单用户模式, ...

  7. js中的event

    event代表事件的状态,例如触发event对象的元素.鼠标的位置及状态.按下的键等等.event对象只在事件发生的过程中才有效.event的某些属性只对特定的事件有意义.比如,fromElement ...

  8. MySQL GTID (四)

    七. GTID的限制以及解决方案 7.1 事务中混合多个存储引擎,会产生多个GTID. 当使用GTID,在同一个事务中,更新包括了非事务引擎(MyISAM)和事务引擎(InnoDB)表的操作,就会导致 ...

  9. 【转】C# 调用 C++ 数据转换

    原文:https://www.cnblogs.com/82767136/articles/2517457.html 在合作开发时,C#时常需要调用C++DLL,当传递参数时时常遇到问题,尤其是传递和返 ...

  10. swarm调度

    Swarm filters Configure the available filters 过滤器分为两类,即节点过滤器和容器配置过滤器. 节点过滤器对Docker主机的特性或Docker守护程序的配 ...