HDU1301 最小生成树 + Prim +（Prim求最小生成树与Dijkstra求最短路的区别） （如此题不要误解题意为用Dijkstra！）

0）

最小生成树，可以用Prim，也可以用Kruskal算法。

Prim（普里姆算法） 与 dijstra 求最短路 很像，但是却不同。

区别是：

前者追求的是，将所有点走一遍，总路径最短，算法是不断更新确定的集合之中的点到集合之外的其他的点的最短距离，全部更新完以后，将这些把所有点连接起来的最短距离加起来就是所求（一定是经过所有点）。

后者追求的是，从起点经过其他所有点，到达终点的路径最短，算法是不断更新从起点到确定的集合之外的其他点的最短距离，全部更新完以后，到终点的最短距离就是起点经过其他点（不一定是经过所有点）到达终点的最短距离。

如下面这个题，

1)Prim算法，是正解

#include <iostream>#include <string.h>#include <stdio.h>#define INF 0x3f3f3f3fusing namespace std;const int maxn=30;int cost[maxn][maxn];int visted[maxn];int dis[maxn];//确定的集合之中的点出发，到未确定的集合之中的最短距离int name(char a){return a-'A'+1;}int Prim(int n){//初始化放入第一个点int cur=1;visted[cur]=1;int sum=0;for(int i=1;i<=n-1;i++){//找出n-1条边for(int j=1;j<=n;j++){if(!visted[j]&&dis[j]>cost[cur][j]){//更新集合之中的点到集合之外的点的最短距离dis[j]=cost[cur][j];}}int minn=INF;int jilu;for(int j=1;j<=n;j++){if(!visted[j]&&minn>dis[j]){//选取各个最短距离之中的最小值，且将到达的点加入集合minn=dis[j];jilu=j;}}visted[jilu]=1;sum+=minn;cur=jilu;}return sum;}int main(){int n;while(~scanf("%d",&n)){if(n==0){break;//!!!??????test}//map <string,int> name;如果用map,那么要遍历mapmemset(cost,INF,sizeof(cost));memset(visted,0,sizeof(visted));memset(dis,INF,sizeof(dis));char a,b;int e_n,v1,v2;for(int i=1;i<=n-1;i++){cin>>a>>e_n;while(e_n--){cin>>b>>v1;cost[name(a)][name(b)]=v1;cost[name(b)][name(a)]=v1;}}cout<<Prim(n)<<endl;}return 0;}

2）dijstra 算出的答案是不对的

3）题目

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

9A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200 

 

Sample Output

21630