POJ 1789-Truck History（最小生成树-不同字母个数）

心已赠人 2022-07-11 13:14 98阅读 0赞

Truck History

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;2000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;26434</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;10259</td> 
  </tr> 
 </tbody> 
</table>

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 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

4
    aaaaaaa
    baaaaaa
    abaaaaa
    aabaaaa
    0

Sample Output

The highest possible quality is 1/3.

Source

[CTU Open 2003][]

## 题目意思： ##

不同的卡车用不同的7位字母表示，每一辆车的编号都可以从其他卡车的编号改变导出而得。

编号从卡车to变成卡车td，改变的质量是d(to,td)。

设计一个导出计划，使得质量1/Σ(to,td)d(to,td)最大，即d(to,td)最小。

##    解题思路： ##

一每辆卡车作为节点，先计算出每辆卡车导出其他卡车编号的质量d(to,td)，将其作为边权值，建立一个无向图。

由题可知是稠密图，所以用prim算法，套模板计算导出边权的和的最小值，即最小生成树。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <algorithm>
    #define MAXN 2010
    #define INF 0xfffffff
    using namespace std;
    int cost[MAXN][MAXN],dis[MAXN],mincost[MAXN],n,t;
    bool used[MAXN];//标识是否使用过
    
    int prim()
    {
        fill(mincost,mincost+n,INF);
        fill(used,used+n,false);
        mincost[0]=0;
        int res=0;//Max=-1;
        while(true)
        {
            int v=-1;
            for(int u=0; u<n; ++u)
            {
                //从不属于已加入生成树的顶点中选取从已加入生成树的点到该顶点的权值最小的点
                if(!used[u]&&(v==-1||mincost[u]<mincost[v]))
                    v=u;
            }
            if(v==-1) break;
            used[v]=true;
            //Max=max(Max,mincost[v]);
            res+=mincost[v];
            for(int u=0; u<n; ++u)
                mincost[u]=min(mincost[u],cost[v][u]);
        }
        return res;
    }
    char s[MAXN][8];
    int number(int i,int j)
    {
        int ans=0;
        for(int k=0; k<7; ++k)
            if(s[i][k]!=s[j][k])//计算两个字符串中不同字母的个数
                ++ans;
        return ans;
    }
    
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        while(cin>>n&&n)
        {
            for(int i=0; i<n; ++i)
                for(int j=0; j<n; ++j)
                    cost[i][j]=INF;
            for(int i=0; i<n; i++)
                cin>>s[i];
            for(int i=0; i<n; i++)
                for(int j=i+1; j<n; ++j)
                {
                    if(cost[i][j]==INF)
                        cost[i][j]=cost[j][i]=number(i,j);//无向图
                    //cout<<cost[i][j]<<endl;
                }
    
            cout<<"The highest possible quality is 1/"<<prim()<<"."<<endl;
        }
        return 0;
    }
    /*
    4
    aaaaaaa
    baaaaaa
    abaaaaa
    aabaaaa
    0
    */

[CTU Open 2003]: http://poj.org/searchproblem?field=source&key=CTU+Open+2003