Jungle Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11855    Accepted Submission(s): 8665

 

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1301

Problem Description

 

Jungle Roads_第1张图片

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.

 

题面翻译

热带岛屿拉格里山的首席长老有一个问题。几年前,大量的外援资金被花在村与村之间的额外道路上。但是丛林无情地超过了道路,所以庞大的公路网维护起来太贵了。长老会必须选择停止维护一些道路。上面左边的地图显示了所有正在使用的道路,以及每个月维护这些道路的成本(AACMS)。当然,在所有的村庄之间都需要有一条道路,即使道路不像以前那么短。首席长老想告诉长老会,他们每月可以在AACMS中花多少钱来维护连接所有村庄的道路。在上面的地图上,村庄被标上了A到I。右边的地图显示了最便宜的维护道路,每月216条AACMS。你的任务是写一个程序来解决这些问题。

 

输入由一到100个数据集组成,后跟一个仅包含0的最后一行。每个数据集以一行开头,该行只包含一个数字n,即1

 

每个数据集的输出为每行一个整数:每个月维护连接所有村庄的道路系统的最低成本(AACMS)。警告:一个暴力解决方案,检查每一个可能的道路集不会完成在一分钟的时间限制。

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

题目意思

给你一个n,然后给出n-1行,表示按字母顺序排列的村庄,每行中有多对字符和数字,表示相连的村庄编号及距离,每条道路的维护费用是道路长度的100倍,要求保证任意两个村庄能直接或间接到达的情况下,每月最小的维护费用。

解题思路

先记录每个村庄之间的道路,然后用Kruskal(克鲁斯卡尔)算法求出最小生成树,就是每月最小的维护费用了。

代码:

#include
#include
#include
#include 
#include 
using namespace std;
int link[27];
typedef struct node{	//自定义结构体
	int cs,bg,ed;
	bool operator < (const node& next)const{
		return cs > next.cs;
	}
}edg;
void init(int n){		//初始化并查集
	for(int i = 0; i <= n; i++){
		link[i] = i;
	}
}
int fd(int point){		//查找该点属于哪个集合
	if(link[point] == point) return point;
	else return fd(link[point]);
}
int main(){
	int n;
	while(~scanf("%d",&n) && n){
		init(n);
		priority_queue q;
		while(!q.empty()) q.pop();
		for(int i = 0; i < n-1; i++){
			char f;
			int num;
			cin>>f>>num;
//			scanf("%c %d",&f,&num);
//			cout<>sf>>jz;
				now.ed = sf - 'A';
				now.cs = jz;
//				cout<

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