杭电 1301 junjle road(最小生成树)

本文探讨了一个关于热带岛屿Lagrishan的难题:如何在有限预算内选择性维护道路,确保所有村庄间的连通性。通过输入数据确定村庄数量、连接关系及维护成本,目标是最小化每月维护成本,同时保持所有村庄间的可达性。
Problem 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. 

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. 
 

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

<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
	char start;
	char end;
	int cost;
 } t[10010];
 int cmp(node a,node b)
 {
 	return a.cost <b.cost ;
 }
 int per[110];
 int find(char c)//定成字符型,比较方便; 
 {
 	char s=c;
 	while(s!=per[s])
 	s=per[s];
 	return s;
 }
 int join(char x,char y)
 {
 	char fx=find(x);
 	char fy=find(y);
 	if(fx!=fy)
 	{
 		per[fx]=fy;
 		return 1;
	 }
	 return 0;
 }
 int m,n,dis;
 char c,ch;
 int main()
{
 	while(scanf("%d",&n),n)
 	{
 		getchar();
 		for(int i=0;i<100;i++)
 		per[i]=i;
 		int js=0;
 		for(int i=1;i<n;i++)
 		{
 		scanf("%c %d",&c,&m);
 		getchar();
 		for(int j=1;j<=m;j++)
 		{
 		scanf("%c %d",&ch,&dis);
 		getchar();
 		t[js].start=c;
 		t[js].end=ch;
 		t[js].cost=dis;
 		js++;
 		}
 		}
 		sort(t,t+js,cmp);
 		int sum=0;
 		for(int i=0;i<js;i++)
 		{
 			if(join(t[i].start,t[i].end))
 			sum=sum+t[i].cost ;
		 }
		 printf("%d\n",sum);
	 }
 return 0;
 }





### 关于最小生成树的 Prim 算法在 PTAC 语言中的实现 PTAC 是一种假设的编程语言,因此其语法和特性需要基于对现有编程语言的理解进行模拟。以下是关于 Prim 算法在 PTAC 语言中的实现或解释。 Prim 算法是一种用于求解图的最小生成树的经典算法,其核心思想是从一个顶点开始逐步扩展已选择的顶点集合,直到覆盖所有顶点[^1]。具体来说,算法始终保持一棵树,并通过不断选择连接当前树与未访问顶点之间的最短边来扩展这棵树。 以下是一个基于伪代码的 Prim 算法在 PTAC 语言中的实现示例: ```ptac function primAlgorithm(graph, startNode): // 初始化数据结构 let visited = array of size graph.size initialized to false let key = array of size graph.size initialized to infinity let parent = array of size graph.size initialized to -1 let priorityQueue = new MinHeap() // 设置起始节点的关键值为 0 key[startNode] = 0 priorityQueue.insert(startNode, 0) while not priorityQueue.isEmpty(): u = priorityQueue.extractMin() visited[u] = true // 遍历相邻节点 for each v in graph.neighbors(u): if not visited[v] and graph.weight(u, v) < key[v]: parent[v] = u key[v] = graph.weight(u, v) priorityQueue.update(v, key[v]) return (parent, key) // 示例调用 let N = input("Enter number of towns: ") let M = input("Enter number of roads: ") let graph = new Graph(N) for i from 1 to M: let u = input("Enter town 1 for road " + i + ": ") let v = input("Enter town 2 for road " + i + ": ") let cost = input("Enter budget cost for road " + i + ": ") graph.addEdge(u, v, cost) let startNode = 1 let result = primAlgorithm(graph, startNode) print("Minimum Spanning Tree edges:") for each node in result.parent: if result.parent[node] != -1: print(result.parent[node], "->", node, ":", result.key[node]) ``` #### 解释 1. **初始化**:`key` 数组记录从当前生成树到每个未访问顶点的最小权重,初始值为无穷大。`parent` 数组记录生成树中每个节点的父节点。 2. **优先队列**:使用最小堆(MinHeap)维护未访问顶点及其对应的 `key` 值,以便快速找到具有最小权重的顶点。 3. **扩展树**:每次从优先队列中取出具有最小 `key` 值的顶点,将其标记为已访问,并更新其邻居节点的 `key` 值。 4. **结果输出**:最终返回生成树的边及其权重。
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