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
- mongodb集群【】
参考 http://www.jianshu.com/p/2825a66d6aed http://www.cnblogs.com/huangxincheng/archive/2012/03/07/238 ...
- jQuery添加删除节点例子第十节"员工增删表"
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/ ...
- MySQL锁学习之UPDATE
##==============================================================================## 学MySQL也蛮长时间了,可一直停 ...
- Ubuntu系统下静态DNS配置详解
1.DNS服务的简介: DNS(Domain Name Server,域名服务器)是进行域名(domain name)和与之相对应的IP地址 (IP address)转换的服务器.DNS中保存了一张域 ...
- JS中有关正则表达式的一些常见应用
总所周知,正则表达式主要用于字符串处理.表单验证等,简单的代码量实现复杂的功能 1.身份证号码的一个校验 先做一个简单的位数校验来判断身份证的合法性:(15位数字或18位数字或17位数字加X|x) v ...
- Get started with Google Analytics
What is Google Analytics Google Analytics is a Google official analytics tool that is primarily used ...
- [转载] 解读ClassLoader
转载自http://www.iteye.com/topic/83978 ClassLoader一个经常出现又让很多人望而却步的词,本文将试图以最浅显易懂的方式来讲解 ClassLoader,希望能对不 ...
- ExtJS+Handler入门显示
<%@ Page Language="C#" AutoEventWireup="true" CodeBehind="Default.aspx.c ...
- 如何管理Session(防止恶意共享账号)——理论篇
目录 知识要求 背景 技术原理 如何管理Session remember me的问题 附录 知识要求 有一定的WEB后端开发基础,熟悉Session的用法,以及与Redis.Database的配合 本 ...
- C++语言中的类型(一)
--分门别类是简化事物最有效的方式. 类型是C++语言的基础,对象类型决定了能对该对象进行的操作. 一.基本内置数据类型 C++预先定义的基本内置数据类型是构造世界万物的原子,数据类型告诉我们数据的意 ...