Jungle Roads
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
Output
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 题目大意:把各个岛屿看成一个点,求各个岛屿之间,权值最小的路径。(最小生成树)
对于数据,数据输入的第一行n代表岛屿的个数,当为0是结束程序,
接着n-1行开始时为这岛屿的编号,用大写字母表示,接着是一个整数m,
表示与该岛屿连接岛屿的个数,然后该行输入m对数据,
第二个数字表示要重修两岛屿之间桥所需要的时间,输出数据见样例及原题。 代码如下:
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cstdlib>
#include <algorithm>
#include <cmath>a
using namespace std;
int p[27];//并查集,用于判断两个点是否直接或间接连通
struct per
{
int u,v,w; }map[80];
bool cmp(per a,per b)
{
return a.w<b.w;
}
int find(int x)
{
return x==p[x]?x:p[x]=find(p[x]);
} int main()
{
int n;
while (scanf("%d",&n),n)
{
int i,j;
for(i=0;i<27;i++)
p[i]=i;
int k=0;
for(i=0;i<n-1;i++)//构造边的信息
{
char str;
int m;
cin>>str>>m;
for(j=0;j<m;j++,k++)
{
char str2;
int t;
cin>>str2>>t;
map[k].u=(str-'A');
map[k].v=(str2-'A');
map[k].w=t;
}
} sort(map,map+k,cmp);//将边从小到大排序
int ans=0;//结果
for(i=0;i<k;i++)
{
int x=find(map[i].u);
int y=find(map[i].v);
if(x!=y)
{//如果两点不在同一连通分量里,则将两点连接,并存储该边 ans+=map[i].w;
p[x]=y;
}
}
printf("%d\n",ans);
}
return 0;
}
Jungle Roads的更多相关文章
- 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 ...
- Jungle Roads[HDU1301]
Jungle Roads Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tota ...
- POJ 1251 Jungle Roads (prim)
D - Jungle Roads Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Su ...
- 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 ...
- POJ1251 Jungle Roads 【最小生成树Prim】
Jungle Roads Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 19536 Accepted: 8970 Des ...
- HDU-1301 Jungle Roads(最小生成树[Prim])
Jungle Roads Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total ...
- Jungle Roads(最小生成树)
Jungle Roads Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...
- POJ 1251 Jungle Roads(最小生成树)
题意 有n个村子 输入n 然后n-1行先输入村子的序号和与该村子相连的村子数t 后面依次输入t组s和tt s为村子序号 tt为与当前村子的距离 求链接全部村子的最短路径 还是裸的最小生成树咯 ...
- Jungle Roads(kruskar)
Jungle Roads 题目链接;http://poj.org/problem?id=1251 Time Limit: 1000MS Memory Limit: 10000K Total Sub ...
- (最小生成树)Jungle Roads -- HDU --1301
链接: http://acm.hdu.edu.cn/showproblem.php?pid=1301 http://acm.hust.edu.cn/vjudge/contest/view.action ...
随机推荐
- ORACLE 创建作业JOB例子
--1.plsql中学习job --学习job --建表 create table test_job(para_date date); commit; insert into test ...
- Sublime Text对Python代码加注释的快捷键
一直在Coursera上补基础课,发现很多课程都用Python作为教学语言,学了一下感觉果然好,简直是用英语在写代码.(我建Python目录的时候发现去年学过一点点Python,居然一点都不记得了= ...
- resignFirstResponder
What is this resignFirstResponder business? Here is the short version:Some view objects are also con ...
- 翻译: TypeScript 1.8 Beta 发布
原文地址:https://blogs.msdn.microsoft.com/typescript/2016/01/28/announcing-typescript-1-8-beta/ 今天,我们发布了 ...
- Hadoop伪分布式配置:CentOS6.5(64)+JDK1.7+hadoop2.7.2
java环境配置 修改环境变量 export JAVA_HOME=/usr/java/jdk1.7.0_79 export PATH=$PATH:$JAVA_HOME/bin export CLASS ...
- python类及其方法
python类及其方法 一.介绍 在 Python 中,面向对象编程主要有两个主题,就是类和类实例类与实例:类与实例相互关联着:类是对象的定义,而实例是"真正的实物",它存放了类中 ...
- 扩展运算是个影藏boss
short a =1; a+=1; //实际上是 a=(short)(a+1); 而 short a=1; a=a+1; //不报错,应为进行算术逻辑运算会默认转为int类型,但是你要把int类型赋值 ...
- HTTP的GET/POST细节
HTTP的GET/POST方式有何区别?这是一个老生常谈的问题,但老生常谈的问题往往有一些让人误解的结论.本文将带您浅尝HTTP协议,在了 解HTTP协议的同时将会展示许多被人们忽视的内容.在掌握了H ...
- linux在yum下安装mysql
1:查看软件是否安装 yum list installed | grep mysql 2:卸载CentOS系统自带mysql数据库 yum -y remove mysql-libs.x86_64,若有 ...
- TListView的一些操作
1,让滚动条滚动的API SetScrollPos int SetScrollPos( _In_ HWND hWnd, _In_ int nBar, _In_ int n ...