标签：c++
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.
题目比较抽象 大概意思就是：
用一个7位的字符串代表一个编号，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。
难点就在于 将抽象的概念转化为图
代码：
#include<iostream>
#include<cstdio>
using namespace std;
#define max 2005
char map[max][20];
int map2[max][max]; //保存每两个字符串的（距离）
int vis[max];
int dis[max];
int value(int x,int y)
{
int s=0;
for(int i=0;i<7;i++)
if(map[x][i]!=map[y][i])
{
s++;
}
return s;
}
int prim(int n)
{
int s=0,pos,i,j,min1;
memset(vis,0,sizeof(vis));
for(i=0;i<n;i++)
dis[i]=map2[0][i]; //假定最小距离就是字符串0到所有其他字符串距离之和
vis[0]=1;
for(i=0;i<n-1;i++)
{
min1=max;
for(j=0;j<n;j++)
{
if(vis[j]==0&&dis[j]<min1) ///先连接一条最短的
{
min1=dis[j];
pos=j;
}
}
vis[pos]=1;
s=s+dis[pos];
for(j=0;j<n;j++)
{
if(vis[j]==0&&dis[j]>map2[pos][j]) /// 查找其他路线是否有 pos到J比0到J短
dis[j]=map2[j][pos];
}
}
return s;
}
int main()
{
int n,i,j;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)
scanf("%s",map[i]);
for(i=0;i<n-1;i++)
for(j=i+1;j<n;j++) //枚举出每两个字符串之间的距离；
map2[i][j]=map2[j][i]=value(i,j);
int ss=prim(n);
printf("The highest possible quality is 1/%d.\n",ss);
}
return 0;
}
poj 1789 Truck History 最小生成树
标签：c++
原文地址：http://blog.csdn.net/axuan_k/article/details/38020715
