POJ 1789 Truck History

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(t_o,t_d)
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

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

4aaaaaaabaaaaaaabaaaaaaabaaaa0

Sample Output

The highest possible quality is 1/3.

Hint:

题目本身比较难看懂。

用一个7位的string代表一个编号，两个编号之间的distance代表这两个编号之间不同字母的个数。一个编号只能由另一个编号“衍生”出来，代价是这两个编号之间相应的distance，现在要找出一个“衍生”方案，使得总代价最小，也就是distance之和最小。
例如有如下4个编号：
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
显然的，第二，第三和第四编号分别从第一编号衍生出来的代价最小，因为第二，第三和第四编号分别与第一编号只有一个字母是不同的，相应的distance都是1，加起来是3。也就是最小代价为3。
问题可以转化为最小代价生成树的问题。因为每两个结点之间都有路径，所以是完全图。
此题的关键是将问题转化为最小生成树的问题。每一个编号为图的一个顶点，顶点与顶点间的编号差即为这条边的权值，题目所要的就是我们求出最小生成树来。这里我用prim算法来求最小生成树。

题目看懂了的话还是比较简单的。

代码如下：

#include<iostream>#include<string>#include<algorithm>#include<vector>#include<iterator>using namespace std;const int MAXV = 2001;const int INF = 0x7fffffff;int V;//结点个数int cost[MAXV][MAXV];bool used[MAXV];int mincost[MAXV];//当前结点距集合中元素的最小值int prim(){    for (int i = 0; i < V; i++)    {        used[i] = false;        mincost[i] = INF;    }    mincost[0] = 0;    int res = 0;    while (true)    {        int v = -1;        for (int i = 0; i < V; i++)        {            if (!used[i] && (v == -1 || mincost[i] < mincost[v]))                v = i;        }        if (v == -1)//所有边均已加入集合中            break;        used[v] = true;//把当前距集合距离的最小顶点加入集合中        res += mincost[v];        for (int i = 0; i < V; i++)        {            mincost[i] = min(mincost[i], cost[v][i]);//更新所有与顶点v有边相边的所有顶点        }    }    return res;}int dis(string& a, string& b){    int t = 0;    for (int i = 0; i < a.size(); i++)    {        if (a[i] != b[i])            t++;    }    return t;}int main(){#ifdef ONLINE_JUDGE#else    freopen("D:\\in.txt", "r", stdin);    freopen("D:\\out.txt", "w", stdout);#endif    int n(0);    vector<string> coll;    string s;    while (cin >> n)    {        if (0 == n)            break;        V = n;        vector<string> coll;        for (int i = 0; i < n; i++)        {            cin >> s;            coll.push_back(s);        }        for (int i = 1; i < coll.size(); i++)        {            for (int j = 0; j < i; j++)                cost[i][j] = cost[j][i] = dis(coll[i], coll[j]);        }        for (int i = 0; i < coll.size(); i++)            cost[i][i] = 0;        int d = prim();        printf("The highest possible quality is 1/%d.\n", d);    }    return 0;}