D - Jungle Roads

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.

Input

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.

Output

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

题意：村子用字母表示，村子间已经有一些公路，求出能确保所有的村子能联通的最小公路和

解法：这就是一个生成最小树的问题，可以用prime算法，不过kru算法更好，前者复杂度为n方，后者为eloge（e为边数）

<span style="font-size:18px;">#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct node
{
    int st,en,len;//存两个村子以及村子间公路长度
}edge[400];//开400够了，因为存的是边数，最多有n*（n-1）/2条边，而题意n小于27
int father[30];//最多才26个村庄，所以30足够了
 int m,n,i,j,k,s,e;
  char ch[2];
bool cmp(node a,node b)
{
    if(a.len<b.len)
        return 1;
    else
        return 0;
}
int find(int a)//判断两个子树是否是同一个祖先
{
    if(a!=father[a])
        return father[a]=find(father[a]);
    return father[a];
}
void kru()
{
    int sum=0;
    for(i=1;i<=j;i++)
    {
        int xx=find(edge[i].st);
        int yy=find(edge[i].en);
        if(xx!=yy)//不是同一祖先说明可以连接这两个村庄
        {
            father[xx]=yy;
            sum+=edge[i].len;//记录权值和
        }
    }
    printf("%d\n",sum);
}
int main()
{
    while(scanf("%d",&n),n)
    {
        for(i=1;i<=n;i++)
            father[i]=i;//节点祖先初始化为其本身
        j=0;//注意这个j代表的是村子间的公路数量，结构体中存的也是公路信息
        for(i=1;i<n;i++)
        {
            scanf("%s%d",&ch,&m);
            s=ch[0]-'A'+1;
            while(m--)
            {
                scanf("%s%d",&ch,&k);
                j++;
                e=ch[0]-'A'+1;
                edge[j].st=s;
                edge[j].en=e;
                edge[j].len=k;
            }
        }
        sort(edge+1,edge+1+j,cmp);//排序很重要，顺序排，保证kru算法的正确性
        kru();
    }
    return 0;
}</span>