博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
POJ 1789:Truck History
阅读量:6583 次
发布时间:2019-06-24

本文共 3306 字,大约阅读时间需要 11 分钟。

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 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

4aaaaaaabaaaaaaabaaaaaaabaaaa0

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 
#include
#include
#include
#include
#include
#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]
>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/"<
<<"."<

版权声明:本文为博主原创文章,未经博主允许不得转载。

转载于:https://www.cnblogs.com/lightspeedsmallson/p/4785784.html

你可能感兴趣的文章
HBase-java api 基本操作
查看>>
POJ2229 Sumsets
查看>>
在LINQ-TO-SQL中实现“级联删除”的方法
查看>>
lemur run PLSA
查看>>
HTTP中的header头解析说明
查看>>
MVC3.0原理学习及总结
查看>>
删除windows中的库、家庭组、收藏夹
查看>>
war 宽度变窄
查看>>
set p4 environment in windows
查看>>
pl/sql development 查询的数据复制到excel
查看>>
自定义指令的参数
查看>>
python实现进度条
查看>>
Android 一个应用启动另一个应用的说明
查看>>
阿里云CentOS7服务器利用LVM分区挂载磁盘全记录
查看>>
Setting up the Web Admin Tool in LDAP 6.x to communicate via SSL
查看>>
SQL好习惯:编写支持可搜索的SQL
查看>>
Shadowbox
查看>>
【 程 序 员 】:伤不起的三十岁,你还有多远 ?
查看>>
openldap安装
查看>>
[leetcode]count and say
查看>>