poj 1251(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

Source

题目链接：http://poj.org/problem?id=1251

一开始感觉是什么高大上的题，后来（⊙ｏ⊙）…原来是最小生成树，如果不会并査集请先去学并査集，那是基础。

这个题题意，给你一个ｎ，代表有ｎ个城市，但是城市用字母表示，接下来ｎ－１行，每一行意思一样，我解释第一组

A 2 B 12 I 25

Ａ表示城市，２表示有两个城市与他相连，Ｂ和Ａ的距离是１２，Ｉ和Ａ的距离是２５

后面一样，求最小生成树

有一点麻烦的地方是把字母变成数字，减去‘Ａ’＋１就ｏｋ啦

直接上代码喽

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define N 30

int father[N];

struct stud{
int a,b,len;
}f[N*N];

int n,k;

int cmp(stud a,stud b)
{
    return a.len<b.len;
}

int cha(int x)
{
    if(x!=father[x])
        father[x]=cha(father[x]);
    return father[x];
}
int main()
{
    int n,i,j,m;
    char c[2];
    int a,len;

    while(scanf("%d",&n),n)
    {
        k=0;
        for(i=1;i<n;i++)
        {
            scanf("%1s%d",c,&m);
            a=c[0]-'A'+1;

            while(m--)
            {
                scanf("%1s%d",c,&len);
                f[k].a=a;
                f[k].b=c[0]-'A'+1;
                f[k].len=len;
                k++;
            }

        }

        sort(f,f+k,cmp);
        for(i=1;i<=n;i++)
            father[i]=i;

        int num=1,ans=0;

        for(i=0;i<k;i++)
        {
            int aa=cha(f[i].a);
            int bb=cha(f[i].b);
            if(aa!=bb)
            {
                father[aa]=bb;
                num++;
                ans+=f[i].len;
                if(num==n)
                    break;
            }
        }

        printf("%d\n",ans);
    }
    return 0;
}

poj 1251(最小生成树)