hdoj-1301-Jungle Roads【最小生成树】
Jungle Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5203 Accepted Submission(s): 3766

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.
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
216
30
pid=1217" target="_blank">
1217
pid=1874" target="_blank">
1874
pid=1598" target="_blank">
1598
pid=2544" target="_blank">
2544
#include<stdio.h>
#include<algorithm>
using namespace std;
struct graph{
int a,b;
int dis;
}G[500];
int root[27];
int find(int i){
if(root[i]==i) return i;
return root[i]=find(root[i]);
}
void unio(int i,int j){
int t=find(i);
int k=find(j);
if(t<=k) root[k]=t;
else root[t]=k;
return;
}
int cmp(graph u,graph v){
return u.dis<v.dis;
}
int kruskal(int n){
sort(G,G+n,cmp);
int shortest=0;
for(int i=0;i<n;++i){
int t=find(G[i].a);
int k=find(G[i].b); if(t!=k){
shortest+=G[i].dis;
unio(t,k);
} }
return shortest;
}
int main(){
int n;
while(~scanf("%d",&n),n){
int i,x,d,u=0,k;
char ch;
for(i=0;i<26;++i) root[i]=i;
for(int p=1;p<n;++p){
getchar();
scanf("%c",&ch); //printf("%c==\n",ch);
x=ch-'A';
scanf("%d",&k);
for(i=1;i<=k;++i){
getchar();
scanf("%c",&ch);//printf("%c$$\n",ch);
scanf("%d",&d);
G[u].a=x,G[u].b=ch-'A';
G[u].dis=d;
++u;
} } printf("%d\n",kruskal(u));
}
return 0;
}
hdoj-1301-Jungle Roads【最小生成树】的更多相关文章
- hdu 1301 Jungle Roads 最小生成树
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1301 The Head Elder of the tropical island of Lagrish ...
- HDOJ 1301 Jungle Roads
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1301 //HDOJ1301 #include<iostream>#include<c ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- hdu Jungle Roads(最小生成树)
Problem Description The Head Elder of the tropical island of Lagrishan has a problem. A burst of for ...
随机推荐
- Graph Cut
转自:http://blog.csdn.net/zouxy09/article/details/8532111 Graph Cut,下一个博文我们再学习下Grab Cut,两者都是基于图论的分割方法. ...
- Python软件源PyPI中国镜像 2016
作为 easy_install 的升级版,pip 为 Pyhton 的包管理提供了极大的方便.一行命令即可完成所需模块的安装: pip install pandas 可是官方镜像的访问速度相当慢,几乎 ...
- sql server2000存储过程sp_droplogin
/* 打开修改系统表的开关 */ sp_configure RECONFIGURE WITH OVERRIDE 存储过程如下: create procedure sp_droplogin @login ...
- 【JBPM4】流程部署
示例代码: ProcessEngine processEngine = Configuration.getProcessEngine(); RepositoryService repositorySe ...
- 跨域请求httpclient
httpclient:是Apache工具包,util,它可以作为一个爬虫,直接爬取某个互联网上的页面.获取到时页面最终的源文件html.直接可以获取页面返回json.就可以直接在代码内部模拟发起htt ...
- Java基础:类加载机制
之前的<java基础:内存模型>当中,我们大体了解了在java当中,不同类型的信息,都存放于java当中哪个部位当中,那么有了对于堆.栈.方法区.的基本理解以后,今天我们来好好剖析一下,j ...
- vue-music 关于Player (播放器组件)--播放和进度条
迷你播放器 1.播放器组件会在各个页面的情况下会打开. 首先在vuex state.js 中定义全局的播放器状态 import {playMode} from 'common/js/config.js ...
- [TCO2013]LitPanels
题意:一个$n\times m$的无色网格,你可以在其中选择两个$x\times y$的子矩形并在其中将其中任意的格子涂上颜色,问最终能得到多少种不同的网格 做这题会用到一个概念叫包围盒(boundi ...
- 【最大流/费用流】BZOJ1834-[ZJOI2010]network 网络扩容
[题目大意] 给定一张有向图,每条边都有一个容量C和一个扩容费用W.这里扩容费用是指将容量扩大1所需的费用.求: 1. 在不扩容的情况下,1到N的最大流: 2. 将1到N的最大流增加K所需的最小扩容费 ...
- 【bfs+优先队列】POJ2049-Finding Nemo
基本上算是普通但略有些繁琐的广搜.给出的墙面和门的坐标为点,而Nemo位于方格中. [思路] 首先思考一下如何存储下整个坐标系.我们预先约定,用一个方格的左下角顶点坐标来作为这个方格的坐标.map[i ...