POJ 1789 Truck History

Posted 2020-09-30 十年换你一句好久不见

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28675		Accepted: 11187

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

4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
0

Sample Output

The highest possible quality is 1/3.

题意：

      给出 n 种卡车，每种卡车的类型是由七个字符组成的，一种卡车可以从另一种卡车派生而来，所需要的代价是两种卡车类型不同的字符个数，求出这 n 种卡车派生的最小代价， n 种车有一 种是开始就已经有的，其余的 n-1 种是派生出来的。

 

//最小生成树
#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath> 
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f 
#define mem(a) (memset(a,0,sizeof(a))) 
typedef long long ll;
int g[2006][2006];
int dis[2006],x,y;
int vis[2006],t,n,m;
char s[2006][10];
int dist(char *a,char *b)
{
    int ans=0;
    for(int i=0;i<7;i++)
    {
        if(a[i]!=b[i]) ans++;
    }
    return ans;
}
int prime()
{
    for(int i=1;i<=n;i++)
    {
        dis[i]=g[1][i];
        vis[i]=0;
    }
    vis[1]=1;
    int minn,sum=0;
    for(int i=1;i<n;i++)
    {
        minn=INF;
        int v;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && minn>dis[j])
            {
                v=j;
                minn=dis[j];
            }
        }
        vis[v]=1;
        if(minn!=INF) sum+=minn;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]) dis[j]=min(dis[j],g[v][j]);
        }
    }
    return sum;
}
int main()
{
    while(scanf("%d",&n) && n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%s",&s[i]);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=i;j++)
            {
                g[i][j]=g[j][i]=dist(s[i],s[j]);
            }
        }
        printf("The highest possible quality is 1/%d.\n",prime());
    }
    return 0;
}