Truck History
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 21376   Accepted: 8311

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.

题目给了N个字符串,每个字符串有7个字符。每个字符串之间都有所谓的“距离”:即不相等的数量。问这些字符串之间一共的距离之和最小是多少。

计算每对字符串之间的距离,之后求其最小生成树。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std; int num;
char truck[2005][10];
int map[2005][2005];
int stack[2005];
int minidis[2005]; int dis(int x,int y)
{
int i,result=0;
for(i=0;i<7;i++)
{
if(truck[x][i]!=truck[y][i])
{
result++;
}
}
return result;
} int prim()
{
int i,j,s,result; memset(stack,0,sizeof(stack));
for(i=1;i<=num;i++)
{
minidis[i]=15;
} stack[1]=1;
minidis[1]=0;
s=1;
result=0;
for(i=1;i<=num-1;i++)
{
int min_all=15;
int min_temp=0;
for(j=2;j<=num;j++)
{
if(stack[j]==0&&minidis[j]>map[s][j])
{
minidis[j]=map[s][j];
}
if(stack[j]==0&&minidis[j]<min_all)
{
min_temp=j;
min_all=minidis[j];
}
}
s=min_temp;
stack[s]=1;
result += min_all;
}
return result;
} int main()
{
int i,j;
while(cin>>num)
{
if(num==0)
break;
for(i=1;i<=num;i++)
{
scanf("%s",truck[i]);
}
for(i=1;i<=num;i++)
{
for(j=i+1;j<=num;j++)
{
map[i][j]=map[j][i]=dis(i,j);
}
}
cout<<"The highest possible quality is 1/"<<prim()<<"."<<endl;
} return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

POJ 1789:Truck History的更多相关文章

  1. POJ 1789:Truck History(prim&amp;&amp;最小生成树)

    id=1789">Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17610   ...

  2. 【POJ 1789】Truck History(最小生成树)

    题意:距离定义为两个字符串的不同字符的位置个数.然后求出最小生成树. #include <algorithm> #include <cstdio> #include <c ...

  3. poj 1789 prime

    链接:Truck History - POJ 1789 - Virtual Judge  https://vjudge.net/problem/POJ-1789 题意:先给出一个n,代表接下来字符串的 ...

  4. POJ 1789 Truck History【最小生成树简单应用】

    链接: http://poj.org/problem?id=1789 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#probl ...

  5. Kuskal/Prim POJ 1789 Truck History

    题目传送门 题意:给出n个长度为7的字符串,一个字符串到另一个的距离为不同的字符数,问所有连通的最小代价是多少 分析:Kuskal/Prim: 先用并查集做,简单好写,然而效率并不高,稠密图应该用Pr ...

  6. POJ 1789 Truck History (Kruskal)

    题目链接:POJ 1789 Description Advanced Cargo Movement, Ltd. uses trucks of different types. Some trucks ...

  7. poj 1789 Truck History

    题目连接 http://poj.org/problem?id=1789 Truck History Description Advanced Cargo Movement, Ltd. uses tru ...

  8. POJ 1789 -- Truck History(Prim)

     POJ 1789 -- Truck History Prim求分母的最小.即求最小生成树 #include<iostream> #include<cstring> #incl ...

  9. poj 1789 Truck History 最小生成树

    点击打开链接 Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15235   Accepted:  ...

随机推荐

  1. 使用Python绘制漫步图

    代码如下: import matplotlib.pyplot as plt from random import choice class RandomWalk(): def __init__(sel ...

  2. DNS原理和智能DNS及CDN

    CDN是什么 CDN(Content Delivery Network)是内容分发网络. 作用: 使用户可以就近取的所需要的内容,加速用户访问网站资源的速度.物理距离更近,响应速度更快.拦截部分网络请 ...

  3. Django——优美的Path()函数

    path( )作用:解析URL地址 path( ) 标准语法: (<>为必须的参数,[]为可选参数) path(<route>, <view>, [name=Non ...

  4. 九十四、SAP中ALV事件之八,显示功能按钮栏

    一.我们把其他代码都注释掉,直接写一行调用 SET PF-STATUS 'TIANPAN_TOOLS'. 二.运行程序,会看到我们上一篇所添加的相关功能栏图标, 三.点击不同图标,会按程序代码,有不同 ...

  5. abstract和interface关键字介绍

    一.abstract关键字介绍 abstract可以修饰方法.类.使用abstract修饰的方法和类分别叫做抽象方法和抽象类. 1.抽象方法 抽象方法的定义:指可以通过abstract关键字声明的方法 ...

  6. poj 1330 Nearest Common Ancestors 求最近祖先节点

    Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 37386   Accept ...

  7. Day 17:缓冲输出字符流和用缓冲输入输出实现登录、装饰者设计模式

    输出字符流 Writer  所有输出字符流的基类,  抽象类. FileWriter 向文件输出字符数据的输出字符流. BufferedWriter 缓冲输出字符流        缓冲输出字符流作用: ...

  8. pytorch随笔

    pytorch中transform函数 一般用Compose把多个步骤整合到一起: 比如说 transforms.Compose([ transforms.CenterCrop(10), transf ...

  9. Hbase PleaseHoldException错误

    PleaseHoldException ① 原因:(由于正在操作Hbase时,电脑突然关机,未正常关闭hbase,故导致shell无法正常显示)如下图: ②解决过程: 先在网上百度到了使用https: ...

  10. 部署Ambari Server实战案例

    部署Ambari Server实战案例 作者:尹正杰 版权声明:原创作品,谢绝转载!否则将追究法律责任. 一.准备三台虚拟机(需要自行安装jdk环境) 1>.角色分配 NameNode节点: h ...