POJ 1789 Truck History（构造最小生成树）

题目链接：http://poj.org/problem?id=1789

Truck History
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 27396 Accepted: 10651
Description

Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks are used for vegetable delivery, other for furniture, or for bricks. The company has its own code describing each type of a truck. The code is simply a string of exactly seven lowercase letters (each letter on each position has a very special meaning but that is unimportant for this task). At the beginning of company’s history, just a single truck type was used but later other types were derived from it, then from the new types another types were derived, and so on.

Today, ACM is rich enough to pay historians to study its history. One thing historians tried to find out is so called derivation plan – i.e. how the truck types were derived. They defined the distance of truck types as the number of positions with different letters in truck type codes. They also assumed that each truck type was derived from exactly one other truck type (except for the first truck type which was not derived from any other type). The quality of a derivation plan was then defined as
1/Σ(to,td)d(to,td)

where the sum goes over all pairs of types in the derivation plan such that to is the original type and td the type derived from it and d(to,td) is the distance of the types.
Since historians failed, you are to write a program to help them. Given the codes of truck types, your program should find the highest possible quality of a derivation plan.
Input

The input consists of several test cases. Each test case begins with a line containing the number of truck types, N, 2 <= N <= 2 000. Each of the following N lines of input contains one truck type code (a string of seven lowercase letters). You may assume that the codes uniquely describe the trucks, i.e., no two of these N lines are the same. The input is terminated with zero at the place of number of truck types.
Output

For each test case, your program should output the text “The highest possible quality is 1/Q.”, where 1/Q is the quality of the best derivation plan.
Sample Input

4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
0
Sample Output

The highest possible quality is 1/3.
Source

CTU Open 2003

【中文题意】给你n个单词，每两个个单词之间的距离为这两个单词中字母不同的个数（即比较每一位，不能对字母排序后再比较），然后由一个字母衍生出其他n-1个字母的花费是多少。
【思路分析】我们根据题意可以把这个题转变一下思维，从某一个点开始，然后可以到达其他n-1个点，这样不是最小生成树么，然后我们可以建边，用Kruskal算法搞一下。
【AC代码】

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;

int pre[2005];

int fin(int x)
{
    if(x==pre[x])
    {
        return x;
    }
    else
    {
        return pre[x]=fin(pre[x]);
    }
}

void join(int x,int y)
{
    if(fin(x)!=fin(y))
    {
        pre[fin(x)]=pre[fin(y)];
    }
}

char a[2005][8];
struct node
{
    int u,v,cost;
}p[4000005];

bool cmp(node x,node y)
{
    return x.cost<y.cost;
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0)break;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",&a[i]);
            pre[i]=i;
        }
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                int s=0;
                for(int k=0;k<7;k++)
                {
                    if(a[i][k]!=a[j][k])
                    {
                        s++;
                    }
                }
                p[cnt].u=i;
                p[cnt].v=j;
                p[cnt++].cost=s;
                //printf("%d\n",p[cnt-1].cost);
            }
        }
        sort(p,p+cnt,cmp);
        int s=0,sum=0;
        for(int i=0;i<cnt;i++)
        {
            if(fin(p[i].u)!=fin(p[i].v))
            {
                join(p[i].u,p[i].v);
                s++;
                sum+=p[i].cost;
            }
            if(s==n-1)break;
        }
        printf("The highest possible quality is 1/%d.\n",sum);
    }
    return 0;
}