最小生成树：POJ1251-Jungle Roads（最小生成树的模板）

POJ 1251 Jungle Roads

        >[poj原址：http://poj.org/problem?id=1251](http://poj.org/problem?id=1251)

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

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解题心得：

• 这个题的题意就是给你一个图，里面有很多的路，要你将这个图变成一棵树，要求树要是所有边的权值加起来最小，也就是最小生成树。既然是树，那么就不可能存在环，所以在生成树的是有有两个注意点，无环且最小。
• 第一个点就是要和最小权值的数，也就是最小生成树了。这个操作可以将所有的边单独的取出来，让这个图全是点然后对边开始排序，在从小的边开始添加到这些点中，那么怎么判断有没有产生环呢，这个就要用到并查集（并查集的详情看：），看两个节点是否有相同的根，如果没有相同的根那么这个边可以添加到里面，如果是相同的根再添加到里面就会产生环。大概就是这么处理的详情可以看看代码。

#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;

int n,t;
int father[28];
struct node//记录一个边的起点终点和长度
{
    int sour;
    int End;
    int len;
} maps[2700];

int find(int a)
{
    if(father[a] == a)
        return a;
    else
        return father[a] = find(father[a]);
}

bool cmp(node a,node b)
{
    return a.len <b.len;
}

void pre_maps()
{
    char now,s;
    int Now;
    t = 0;
    for(int i=1; i<n; i++)
    {
        scanf(" %c",&s);
        scanf("%d",&Now);
        while(Now--)
        {
            int len;
            scanf(" %c",&now);
            scanf("%d",&len);
            maps[t].sour = s - 'A';
            maps[t].End = now - 'A';
            maps[t].len = len;
            t++;
        }
    }
}

void merge(int a,int b)
{
    int fa = find(a);
    int fb = find(b);
    if(fa != fb)
        father[fb] = fa;
}

int main()
{
    while(scanf("%d",&n) && n)
    {
        for(int i=0; i<28; i++)//将并查集的father初始化
            father[i] = i;
        int ans = 0;
        pre_maps();
        sort(maps,maps+t,cmp);//将边从小到大拍一个序，排序要用边长作为标准
        for(int i=0; i<t; i++)
        {
            int a,b;
            a = find(maps[i].sour);
            b = find(maps[i].End);
            if(a != b)//两个点不同根则可以添加边
            {
                ans +=  maps[i].len;
                merge(maps[i].sour,maps[i].End);//添加了边之后将这两个点合并
            }
        }
        printf("%d\n",ans);
    }
}