POJ_1789_最小生成树

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.

解题思路

最小生成树：一棵树的形式，从头到尾，一条线连接。无向图（即1-2与2-1距离一样）。
首先任意选一个顶点加入生成树。接下来找一条边加入到生成树，枚举每一个树顶点到每一个非树顶点所有的边，找到最短边加入到生成树，重复n-1次，直到所有的顶点都加入生成树

• dis[]中存储的是每个顶点到生成树的距离，而不是到源点的距离

模板代码：

 prim算法://进行n-1次，将n个顶点都加入到生成树

 book[1]=1;//先将1号顶点放入生成树，book数组用来记录顶点是否已经加入生成树
 count++;//
 while(count<n)
 {
     min=inf;//inf为无穷大
     //选择最近的点 dis[]
     for(i=1;i<=n;i++)
     {
         if(book[i]==0&&dis[i]<min)
         {
             min=dis[i];
             j=i;
         }
     }
     count++;
     book[j]=1;
     //以j为当前顶点，遍历与j相连的所有的边，以j为中间点，更新生成树到每一个非树顶点的距离。注意：dis[]中存储的是每个顶点到生成树的距离，而不是到源点的距离
     for(k=1;k<=n;k++)
     {
         if(book[k]==0&&dis[k]>e[j][k])
             dis[k]=e[j][k];
     }
     sum=sum+dis[j];//sum=sum+min;将所有的最小边相加，即为最后生成树的距离。
 }

本题思路：
题意：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。

转化成最小生成树问题：首先将每一行到每一行的“距离”计算出来，问题就解决了，每行代表顶点，两两之间不同的个数即为距离，进而转化成矩阵，利用prim算法即可求解。

代码

#include <stdio.h>
int e[2005][2005];
int n;
void prim()
{
    int i,j,k,min,inf=99999999;
    int book[2005]={0};
    int dis[2005];
    int count=0,result=0;
    for(i=0;i<n;i++)
        dis[i]=e[0][i];
    book[0]=1;
    count++;
    while(count<n)
    {
        min=inf;
        for(i=0;i<n;i++)
        {
            if(book[i]==0&&dis[i]<min)
            {
                min=dis[i];
                j=i;
            }
        }
        book[j]=1;
        count++;

        //以当前j（以确定为树的顶点）为中间点，更新生成树到每一个非树顶点的距离
        for(k=0;k<n;k++)
        {
            if(book[k]==0&&dis[k]>e[j][k])
                dis[k]=e[j][k];
        }
        result=result+dis[j];//已确定的最短距离
    }
    printf("The highest possible quality is 1/%d.\n",result);
}
int main()
{
    int i,j,k,sum;
    char s[2005][2005];
    while(scanf("%d",&n)!=EOF&&n)
    {
        for(i=0;i<n;i++)
            scanf("%s",s[i]);
        for(i=0;i<n;i++)
        {
            for( j=0;j<n;j++)
            {
                sum=0;
                for(k=0;k<7;k++)
                {
                    if(s[i][k]!=s[j][k])//任一两行不相等的
                        sum++;
                }
                e[i][j]=e[j][i]=sum;//距离，转化成矩阵形式

            }
        }
        prim();
        /*for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
                printf("%d ",e[i][j]);
            printf("\n");
        }*/


    }
    return 0;
}