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
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=30,M=1000;
int par[N],n,m,ans;
struct edge{int u,v,w;} e[M];
bool cmp(edge a,edge b) {return a.w<b.w;} int Find(int x)
{
int r=x,tmp;
while(par[r]>=0) r=par[r];
while(x!=r)
{
tmp=par[x];
par[x]=r;
x=tmp;
}
return r;
} void Union (int u,int v)
{
int ru=Find(u),rv=Find(v),tmp=par[ru]+par[rv];
if(par[ru]<par[rv])
par[rv]=ru,par[ru]=tmp;
else
par[ru]=rv,par[rv]=tmp;
} void kruskal()
{
memset(par,-1,sizeof(par));
int cnt=0;
for(int i=1;i<=m;++i)
{
int u=e[i].u,v=e[i].v;
if(Find(u)!=Find(v))
{
++cnt;
ans+=e[i].w;
Union(u,v);
}
if(cnt>=n-1) break;
}
} int main()
{
char s[2]; int t,tt;
while(scanf("%d",&n),n)
{
m=0;
for(int i=1;i<n;++i)
{
scanf("%s%d",s,&t);
for(int j=1;j<=t;++j)
{
scanf("%s%d",s,&tt);
e[++m].u=i,e[m].v=s[0]-'A'+1,e[m].w=tt;
}
} sort(e+1,e+m+1,cmp);
ans=0; kruskal();
printf("%d\n",ans);
}
return 0;
}
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
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
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
POJ 1251 Jungle Roads(最小生成树)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 ...
- POJ 1251 Jungle Roads - C语言 - Kruskal算法
Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid ...
- POJ 1251 Jungle Roads (prim)
D - Jungle Roads Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Su ...
- POJ 1251 Jungle Roads (最小生成树)
题目: Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign ...
- 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 ...
- [ An Ac a Day ^_^ ] [kuangbin带你飞]专题六 最小生成树 POJ 1251 Jungle Roads
题意: 有n个点 每个点上有一些道路 求最小生成树 解释下输入格式 A n v1 w1 v2 w2 A点上有n条边 A到v1权值是w1 A到v2权值是w2 思路: 字符串处理之后跑kruskal求最小 ...
- POJ - 1251 Jungle Roads (最小生成树&并查集
#include<iostream> #include<algorithm> using namespace std; ,tot=; const int N = 1e5; ]; ...
- POJ 1251 Jungle Roads(Kruskal算法求解MST)
题目: The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money w ...
- POJ 1251 Jungle Roads (zoj 1406) MST
传送门: http://poj.org/problem?id=1251 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=406 P ...
随机推荐
- Spring各jar包作用及依赖
先附spring各版本jar包下载链接http://repo.spring.io/release/org/springframework/spring/ spring.jar 是包含有完整发布模块的单 ...
- Java工具类之——BigDecimal运算封装(包含金额的计算方式)
日常对于金额计算,应该都是用的BigDecimal, 可是苦于没有好的工具类方法,现在贡献一个我正在用的对于数字计算的工具类,项目中就是用的这个,简单粗暴好用,话不多说,代码奉上(该工具类需要引入g ...
- airodump-ng使用手册
选项: -i, --ivs 捕捉WEP加密的包,忽略出IV之外的所有的包,保存为.ivs格式 airodump-ng wls35u1 -i -w captures airodump-ng wls35u ...
- MUI框架开发HTML5手机APP(一)--搭建第一个手机APP
前 言 JRedu 随着HTML5的不断发展,移动开发成为主流趋势!越来越多的公司开始选择使用HTML5开发手机APP,而随着手机硬件设备配置的不断提升,各种开发框架的不断优化,也使着H5开发的 ...
- 引用reference作用域scope闭包closure上下文context用法
引用(reference).作用域(scope).闭包(closure)以及上下文(context)是JavaScript重中之重的基础,也是学习好JavaScript的基础.在这里我以浅显的理解给大 ...
- Snow and Rainbow
缘分,让我们走到了一起.让这个美好的时刻美好的回忆记录在这里吧.
- 用StringBuilder和StringBuffer实现的Unicode解码方法的比较(Java)
初衷是用正则来写一个Unicode字符串转码的方法,一开始是打算结合StringBuilder写的,但是看到jdk7的Matcher.appendReplacement文档中一段示例代码用了Match ...
- linux平台搭建postfix邮件服务器
一,搭建邮件服务器前准备如下: Centos 7.2 64位Postfix-2.8.12.tar.gz Postfix MTA(邮件传输代理)Dovecot-2.1.8.tar.gz IMAP 和 P ...
- sql 触发器,看完后对CHK有更深的理解
触发器是一种特殊类型的存储过程,它不同于之前的我们介绍的存储过程.触发器主要是通过事件进行触发被自动调用执行的.而存储过程可以通过存储过程的名称被调用. 什么是触发器? 触发器对表进行插入.更新.删除 ...
- SQLServer2008数据库连接error40错误
在连接SQL Server偶尔会遇到报错,如在与 SQL Server 建立连接时出现与网络相关的或特定于实例的错误.未找到或无法访问服务器.请验证实例名称是否正确并且 SQL Server 已配置为 ...