Truck History(Prim 最小生成树)

Truck History
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 29200 Accepted: 11417
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.

举例：
abaaaba 以第一行为基准与各行的差值为 0 2 6 1
baaaaba 以第二行为基准与各行的差值为 2 0 4 3
baabbab 以第三行为基准与各行的差值为 6 4 0 5
abaaaaa 以第四行为基准与各行的差值为 1 3 5 0
邻接矩阵，加权无向图~

7 个小写字母来表示每种truck的型号，每两种型号之间的差距为字母串中不同字母的个数。现在给出 n 种不同型号的truck，问怎样使 1/∑(to,td)d(to,td) 的值最小。

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;char ma[2002][2002];//存图int mma[2002][2002];//邻接矩阵，差值int n;int dist[12091];int v[12901];void Prim()//最小生成树{    int i;    memset(v, 0, sizeof(v));//初始化    for(int i=1;i<=n;i++)//得到dist数组        dist[i] = mma[1][i];    v[1] = 1;    int min;    int point;    int ans = 0;    for(i=2;i<=n;i++)    {        point = i;        min = INF;        for(int j=1;j<=n;j++)        {            if(v[j]==0&&dist[j]<min)            {                min = dist[j];                point = j;            }        }        ans += min;        v[point] = 1;        for(int j=1;j<=n;j++)        {            if(mma[point][j]<dist[j]&&v[j]==0)                dist[j] = mma[point][j];        }    }    cout<<"The highest possible quality is 1/"<<ans<<'.'<<endl;//输出}int main(){    while(cin>>n&&n)    {        getchar();        for(int i=1;i<=n;i++)//输入        {            cin>>ma[i];        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                int ans =0;                for(int k=0;k<7;k++)                {                    if(ma[i][k]!=ma[j][k])//求出每两行之间的不同字母的个数                        ans++;                }                mma[i][j] = ans;//不同字母的个数为邻接矩阵的各个值            }        }        Prim();    }    return 0;}