Truck History - POJ 1789 最小生成树

最新推荐文章于 2017-08-17 16:36:54 发布

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针对一组卡车类型代码，本篇博客介绍了一种算法来找出这些类型间最优的衍生路径。通过计算不同类型的卡车代码之间的距离，并利用最小生成树算法，最终确定了连接所有卡车类型的最短距离总和。

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17900		Accepted: 6915

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_{(t_o,t_d)}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：两个字符串有多少不同的，就是说他们的距离是多少。然后问把他们都连起来最短的距离和是多少。

思路：直接最小生成树模板。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[2005][10];
struct node
{ int u,v,w;
}edge[4000005];
bool cmp(node a,node b)
{ return a.w<b.w;
}
int num,ans,n,p[2005];
int f(int x)
{ return x==p[x] ? x : p[x]=f(p[x]);
}
void Kruskal()
{ int u,v,i;
  for(i=1;i<=n;i++)
   p[i]=i;
  for(i=1;i<=num;i++)
  { u=f(edge[i].u);
    v=f(edge[i].v);
    if(u!=v)
    { p[v]=u;
      ans+=edge[i].w;
    }
  }
}
int main()
{ int i,j,k,p;
  while(~scanf("%d",&n) && n)
  { for(i=1;i<=n;i++)
     scanf("%s",s[i]);
    num=0;
    for(i=1;i<=n;i++)
     for(j=i+1;j<=n;j++)
     { p=0;
       for(k=0;k<7;k++)
        if(s[i][k]!=s[j][k])
         p++;
       edge[++num].u=i;
       edge[num].v=j;
       edge[num].w=p;
     }
    sort(edge+1,edge+1+num,cmp);
    ans=0;
    Kruskal();
    printf("The highest possible quality is 1/%d.\n",ans);
  }
}