POJ 1789 Truck History

我只能说，题目描述的好像蛮厉害的样子~

其实还只是一个prim的模板题~Orz

题意：给你N个字符串，每个字符串都只有7位，每两个字符串会有一个“距离”，所谓的距离就是从0号位---6号位，一一对应过去，有1个字符不一样就+1.所以距离就是两个字符串的距离就是【0,7】，这样就可以把每两个字符串的距离算出来，而且第一个字符串不是派生出来的，所以就可以从第一个走起，建一个最小生成树。

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14101		Accepted: 5366

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.

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
const int M = 2005;
int n;
char name[M][10];
bool visit[M];
int mp[M][M];
int dis[M];

void prim( int st )
{
for( int i=0 ; i<n ; i++ ){
visit[i] = false;
dis[i] = mp[st][i];
}
dis[st] = 0;
visit[st] = true;
int ans = 0;
int tmp , k;
for( int i=0 ; i<n ; i++ ){
tmp = 99999999;
for( int j=0 ; j<n ; j++ ){
if( !visit[j] && tmp > dis[j] )
tmp = dis[ k = j ];
}
if( tmp == 99999999 )
break;
ans += tmp;
visit[k] = true;
for( int j=0 ; j<n ; j++ ){
if( !visit[j] && dis[j] > mp[k][j] )
dis[j] = mp[k][j];
}
}
printf("The highest possible quality is 1/%d.\n",ans);
}

int main()
{
while( scanf("%d",&n) == 1 && n ){
for( int i=0 ; i<n ; i++ )
scanf("%s",name[i]);
for( int i=0 ; i<n ; i++ ){
for( int j=i+1 ; j<n ; j++ ){
mp[i][j] = 0;
for( int k=0 ; k<7 ; k++ ){
if( name[i][k] != name[j][k] )
mp[i][j]++;
}
mp[j][i] = mp[i][j];
}
}
prim( 0 );
}
return 0;
}