poj1251 Jungle Roads

題目題目題目：

2:Jungle Roads
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總時間限制: 1000ms 內存限制: 65536kB
描述

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.

輸入
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.
輸出
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.
樣例輸入
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
樣例輸出
216
30
來源
Mid-Central USA 2002

用Prim算法求最小生成樹(Minimum-cost Spanning Tree, MST)。可以看出Prim算法和Dijkstra算法在框架上是相同的，唯一的區別就是：在抽取出權值最小的邊之後，Dijkstra算法對這個邊進行鬆弛操作，Prim算法直接將這個邊納入MST，然後都將這條邊的所有未訪問鄰接邊儲存起來備用。

代碼清單：

//Prim算法生成Minimum-cost Spanning Tree(MST)
//圖的儲存方式還是用我目前熟悉的鄰接矩陣
#include <iostream>
#include <string>
using namespace std;

#define INFINITE 1000000
#define MAXN 26
#define MAXR 75	//最大路徑數目

typedef struct _edge
{
	int begin_vertex;
	int end_vertex;
	int length;
} edge;

int dist[MAXN][MAXN];
int visited[MAXN];
edge edge_store[MAXR];

void init(int n)
{
	//初始化鄰接矩陣
	for (int i=0; i<n; ++i)
	{
		for (int j=0; j<n; ++j)
		{
			dist[i][j]=INFINITE;	//孤立的點，沒有通路
		}
	}
	for (int i=0; i<n; ++i)
	{
		dist[i][i]=0;	//自己到自己的距離是0
	}

	//初始化visited數組
	for (int i=0; i<n; ++i)
	{
		visited[i]=0;	//0沒被訪問，1已被訪問
	}

	//初始化edge數組
	for (int i=0; i<MAXR; ++i)
	{
		edge_store[i].begin_vertex=-1;
		edge_store[i].end_vertex=-1;
		edge_store[i].length=INFINITE;
	}
}

void build_graph(int n)
{
	string str;
	char str_c;
	int to_count, from, to, d;

	for (int i=0; i<n-1; ++i)
	{
		cin>>str;
		cin>>to_count;

		str_c=str[0];
		from=str_c-'A';

		for (int j=0; j<to_count; ++j)
		{
			cin>>str;
			cin>>d;

			str_c=str[0];
			to=str_c-'A';

			dist[from][to]=d;
			dist[to][from]=d;
		}
	}
}

//prim算法核心代碼
int extract_min(int size)	//抽取長度最短的邊
{
	int min_len=INFINITE;
	int v=-1;
	for (int i=0; i<size; ++i)
	{
		if (visited[edge_store[i].end_vertex]==0 && edge_store[i].length<min_len)	//沒有訪問過且邊的長度更短
		{
			min_len=edge_store[i].length;
			v=i;
		}
	}
	return v;
}

int run_prim(int n)
{
	int mtw=0;	//最小權值總和
	visited[0]=1;	//始終從A點開始迭代
	int k=0, e;

	int current_from=0;
	for (int i=0; i<n-1; ++i)	//要添加(n-1)條邊
	{
		for (int j=0; j<n; ++j)	//儲存當前頂點的所有鄰接邊
		{
			if (dist[current_from][j]<INFINITE && visited[j]==0)	//連通且沒有訪問的鄰接邊
			{
				edge_store[k].begin_vertex=current_from;
				edge_store[k].end_vertex=j;
				edge_store[k].length=dist[current_from][j];
				++k;
			}
		}

		e=extract_min(k);	//取出長度最小的鄰接邊在數組中的下標
		mtw+=edge_store[e].length;	//累加到路徑總長中
		visited[edge_store[e].end_vertex]=1;	//邊的終點標記爲已訪問

		current_from=edge_store[e].end_vertex;	//下一個起點就是當前的終點
	}

	return mtw;
}

int main()
{
	//freopen("D:\\in.txt", "r", stdin);  
	//freopen("D:\\out.txt", "w", stdout);  

	int n, mtw;
	
	while (1)
	{
		cin>>n;
		if (n==0) break;

		init(n);
		build_graph(n);
		mtw=run_prim(n);

		cout<<mtw<<endl;
	}

	return 0;
}