zoj 2158 Truck History（最小生成树））

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 

where the sum goes over all pairs of types in the derivation plan such that t_o is the original type and t_d the type derived from it and d(t_o,t_d) 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 Specification

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 Specification

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

4aaaaaaabaaaaaaabaaaaaaabaaaa0

Sample Output

The highest possible quality is 1/3.

题意：有n辆卡车，卡车的编号由7为小写字母的字符串组成，优劣值的计算方法是：将卡车类型编号的距离定义成卡车类型编码字符串中（7个位置上）不同字符串的位置数目。

要使优劣值最大，那么其距离和应该最小，把每个字符串看成一个顶点，距离和看成边长，那么就是求最小生成树。

#include <iostream>#include <algorithm>using namespace std;int Edge[2010][2010];int lowcost[2010];int n;char num[2010][10];int prim(){    int sum=0;    for (int i=0; i<n; i++)    {        lowcost[i]=Edge[0][i];    }    lowcost[0]=-1;    int k=0;    for (int i=1; i<n; i++)    {        int min=100000;        for (int j=0; j<n; j++)        {            if(lowcost[j]!=-1 && lowcost[j]<min)            {                min=lowcost[j];                k=j;            }        }        sum+=min;        lowcost[k]=-1;        for (int j=0;j<n; j++)        {            if(Edge[k][j]<lowcost[j])            {                lowcost[j]=Edge[k][j];            }        }    }    return sum;}int main(){    while (cin>>n && n)    {        for (int i=0; i<n; i++)        {            cin>>num[i];        }        for (int i=0; i<n; i++)        {            for (int j=i+1; j<n; j++)            {                Edge[i][j]=0;                int ans=0;    //权值                for (int k=0; k<7; k++)                {                    if(num[i][k]!=num[j][k])                    {                        ans++;                    }                }                Edge[i][j]=Edge[j][i]=ans;               }        }        cout<<"The highest possible quality is 1/"<<prim()<<"."<<endl;    }    return 0;}