POJ-1251: Jungle Roads（并查集+kruskal）

最小生成树

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：首先输入n，当n输入为0时程序结束。接下来的n-1行，先输入起点，A~Z大写字母表示，之后输入终点个数，然后输入终点（也是A~Z表示）和到终点所需要的距离。

• 并查集： 判断两个元素是否属于同一集合。初始化的时候，一个集合的father（父节点）都是这个集合自己的标号。没有跟它同类的集合，那么这个集合的源头只能是自己了，因此使father[i]=i。
  如果集合的father等于集合的编号（即还没有被合并或者没有同类），那么自然返回自身编号。
  如果不同，那么就可以调用递归函数。

int findx(int x)
{
    if(x!=father[x]) //如果父节点不同
        father[x]=findx(father[x]); //调用递归
    return father[x];
}

• kruskal:求加权连通图的最小生成树的算法。先将所有的边排序。kruskal算法总共选择n-1条边，（共n个点）所使用的贪婪准则是：从剩下的边中选择一条不会产生环路的具有最小耗费的边加入已选择的边的集合中。当考虑某条边时，若将其加入到已选边的集合中会出现环路，则将其抛弃，否则，将它选入。使用并查集判断是否产生回路。若不会产生回路，就合并他们。

void kruskal()
{
    int i,x,y;
    sum=0;
    for(i=1;i<=11111;i++) //首先初始化father，每个节点的父节点都是其本身
        father[i]=i;
    for(i=0;i<k;i++)
    {
        x=findx(a[i].st); //并查集判断父节点是否相同
        y=findx(a[i].en);
        if(x!=y) //若父节点不同，不会产生回路
        {
            sum+=a[i].len; //将该边加入集合
            father[x]=y; //合并他们的父节点
        }
    }
}

• 完整代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int k,sum,father[11111];
struct node
{
    int st,en,len;
}a[11111];
int cmp(node a,node b)
{
    return a.len<b.len;
}
int findx(int x)
{
    if(x!=father[x])
        father[x]=findx(father[x]);
    return father[x];
}
void kruskal()
{
    int i,x,y;
    sum=0;
    for(i=1;i<=11111;i++)
        father[i]=i;
    for(i=0;i<k;i++)
    {
        x=findx(a[i].st);
        y=findx(a[i].en);
        if(x!=y)
        {
            sum+=a[i].len;
            father[x]=y;
        }
    }
}
int main()
{
    int n,i,m,x;
    char ch1[3],ch2[3];
    while(scanf("%d",&n),n)
    {
        memset(father,0,sizeof(father));
        k=0;
        for(i=0;i<n-1;i++)
        {
            scanf("%s%d",ch1,&m);
            while(m--)
            {
                scanf("%s%d",ch2,&x);
                a[k].st=ch1[0]-'A';
                a[k].en=ch2[0]-'A';
                a[k].len=x;
                k++;
            }
        }
        sort(a,a+k,cmp);
        kruskal();
        printf("%d\n",sum);
    }
    return 0;
}