poj1789

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14921		Accepted: 5703

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.

最小生成树

#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;

const int inf=10;          //无穷大（两点间边权最大为7）
const int large=2001;

int n;  //truck types
char str[large][8];
int dist[large][large]={0};

/*Compute Weight*/

int weight(int i,int j)     //返回两个字符串中不同字符的个数（返回边权）
{
	int w=0;
	for(int k=0;k<7;k++)
		if(str[i][k]!=str[j][k])
			w++;
	return w;
}

/*Prim Algorithm*/

int prim(void)
{
	int s=1;       //源点（最初的源点为1）
	int m=1;       //记录最小生成树的顶点数
	bool u[large]; //记录某顶点是否属于最小生成树
	int prim_w=0;  //最小生成树的总权值
	int min_w;     //每个新源点到其它点的最短路
	int flag_point;
	int low_dis[large];  //各个源点到其它点的最短路

	memset(low_dis,inf,sizeof(low_dis));
	memset(u,false,sizeof(u));
	u[s]=true;

	while(1)
	{
		if(m==n)      //当最小生成树的顶点数等于原图的顶点数时，说明最小生成树查找完毕
			break;

		min_w=inf;
		for(int j=2;j<=n;j++)
		{
			if(!u[j] && low_dis[j]>dist[s][j])
				low_dis[j] = dist[s][j];
			if(!u[j] && min_w>low_dis[j])
			{
				min_w=low_dis[j];
				flag_point=j;      //记录最小权边中不属于最小生成树的点j
			}
		}
		s=flag_point;       //顶点j与旧源点合并
		u[s]=true;          //j点并入最小生成树（相当于从图上删除j点，让新源点接替所有j点具备的特征）
		prim_w+=min_w;      //当前最小生成树的总权值
		m++;                
	}
	return prim_w;
}

int main(void)
{
	int i,j;

	while(cin>>n && n)
	{
		/*Input*/
		
		for(i=1;i<=n;i++)
			cin>>str[i];

		/*Structure Maps*/

		for(i=1;i<=n-1;i++)
			for(j=i+1;j<=n;j++)
				dist[i][j]=dist[j][i]=weight(i,j);

		/*Prim Algorithm & Output*/

		cout<<"The highest possible quality is 1/"<<prim()<<'.'<<endl;

	}
	return 0;
}