poj 1789 Truck History【最小生成树prime】

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21518		Accepted: 8367

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.

我也是挺纳闷 为什么我把 建图和prime算法过程 写作子函数放在main函数外边怎么就不能调用呢？
题意：给出卡车的编号，编号是一个长度为7的字符串，一个编号由另一个编号衍生出来，而衍生出来所需要的代价就是两个编号之间不同字符的个数
     （相当于两个顶点相连后的权值），现在求给出的这组字符串所需要的最小代价

#include<stdio.h>
#include<string.h>
#define MAX 2010
#define INF 0x3ffffff
int map[MAX][MAX];
char s[MAX][11];
int low[MAX],vis[MAX];
int t;
int fun(int i,int j)
{
	int ans=0;
	for(int k=0;k<7;k++)
	{
		if(s[i][k]!=s[j][k])
		    ans++;
	}
	return ans;
}
void init()
{
	int i,j;
	for(i=0;i<t;i++)
	{
		for(j=0;j<t;j++)
		{
			if(i==j)
			map[i][j]=0;
			else
			map[i][j]=INF;
		}
	}
}
int main()
{
	int t;
	int i,j;
	while(scanf("%d",&t),t)
	{
		init();
		for(i=0;i<t;i++)
		scanf("%s",s[i]);
		for(i=0;i<t-1;i++)
		{
			for(j=i+1;j<t;j++)
			{
				map[i][j]=map[j][i]=fun(i,j);
			}
		}
	    int next,min,mindis=0;
	    memset(vis,0,sizeof(vis));
	    for(i=0;i<t;i++)
	    low[i]=map[0][i];
	    vis[0]=1;
	    for(i=0;i<t-1;i++)
	    {
	        min=INF;
	        for(j=0;j<t;j++)
	        {
	            if(!vis[j]&&min>low[j])
	            {
	                min=low[j];
	                next=j;
	            }
	        }
	        mindis+=min;
	        vis[next]=1;
	        for(j=0;j<t;j++)
	        {
	            if(!vis[j]&&low[j]>map[next][j])
	            low[j]=map[next][j];
	        }
	    }
	    printf("The highest possible quality is 1/%d.\n",mindis);
	}
	return 0;
}