Truck History
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 20768   Accepted: 8045

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.

代码
基本prim模板没什么可说的,只是需要将字符串预处理为邻接矩阵即可

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int map[2005][2005];
char str[2005][7];
int vis[2005],dis[2005];
int n;
int prim(int u){
int sum=0;
for(int i=1;i<=n;i++){
dis[i]=map[u][i];
}
vis[u]=1;
for(int i=1;i<n;i++){
int tmin=999999999;
int ans;
for(int j=1;j<=n;j++){
if(dis[j]<tmin&&!vis[j]){
tmin=dis[j];
ans=j;
}
}
sum+=tmin;
vis[ans]=1;
for(int k=1;k<=n;k++){
if(dis[k]>map[ans][k]&&!vis[k])
dis[k]=map[ans][k];
}
}
return sum;
}
int main(){

while(scanf("%d",&n)!=EOF){
if(n==0)
break;

memset(map,0,sizeof(map));
memset(str,0,sizeof(str));
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));
getchar();
for(int i=1;i<=n;i++){
scanf("%s",str[i]);
getchar();

}
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
int sum=0;
for(int k=0;k<7;k++){
if(str[i][k]!=str[j][k])
sum++;
}
map[i][j]=sum;
map[j][i]=sum;
}
}
printf("The highest possible quality is 1/%d.\n",prim(1));
}
return 0;
}

poj1789 Truck History的更多相关文章

  1. POJ1789 Truck History 【最小生成树Prim】

    Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18981   Accepted: 7321 De ...

  2. POJ1789 Truck History 2017-04-13 12:02 33人阅读 评论(0) 收藏

    Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 27335   Accepted: 10634 D ...

  3. poj1789 Truck History最小生成树

    Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 20768   Accepted: 8045 De ...

  4. POJ1789:Truck History(Prim算法)

    http://poj.org/problem?id=1789 Description Advanced Cargo Movement, Ltd. uses trucks of different ty ...

  5. POJ1789 Truck History(prim)

    题目链接. 分析: 最大的敌人果然不是别人,就是她(英语). 每种代表车型的串,他们的distance就是串中不同字符的个数,要求算出所有串的distance's 最小 sum. AC代码如下: #i ...

  6. Truck History(prim & mst)

    Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19772   Accepted: 7633 De ...

  7. 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:  ...

  8. poj 1789 Truck History

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

  9. POJ 1789 Truck History (最小生成树)

    Truck History 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/E Description Advanced Carg ...

随机推荐

  1. DOM(九)使用DOM设置文本框

    1.控制用户输入的字符个数 对于单行文本框和密码输入框,可以利用maxlength属性控制用户输入的字符个数. 对于多行文本,maxlength为自定义属性,其值最多输入的字符的个数,在onkeypr ...

  2. Intelli IDEA 14.04开发版+注册码

    Idea,一款比较出色的IDE开发工具,懂她的人,自然懂得. 本版本-14.04+注册码 注册界面  注册码:               Name:happy                KEY: ...

  3. Source Tree for MAC1.6

    Atlassian ID has become the new Atlassian Account. Read more about it here. After some great communi ...

  4. Customizing Navigation Bar and Status Bar

    Like many of you, I have been very busy upgrading my apps to make them fit for iOS 7. The latest ver ...

  5. Android平台的开发环境的发展演变

    因为之前学习java语言的时候安装过了eclipse,所以想在eclipse上搭建android平台,在参照知乎上大神们的意见,发现了AS强大的代码提示.实时预览和搜索匹配等出色功能,最后还是选择在A ...

  6. QLabel添加图片链接

    <a href = "http://1450591036.qzone.qq.com"><img src=://splash.png height="30 ...

  7. phpstudy2016最新版本mysql无法使用innodb的问题解决

    这里顺便记录一下今天遇见的神奇问题,在使用官方最新版本的phpstudy中,其它组件一切正常,但是奇怪的发现mysql是无法开启innodb的,以下为最新的下载地址: http://www.phpst ...

  8. poj 2891 扩展欧几里得迭代解同余方程组

    Reference: http://www.cnblogs.com/ka200812/archive/2011/09/02/2164404.html 之前说过中国剩余定理传统解法的条件是m[i]两两互 ...

  9. C++处理一个动态规划的问题

    嗯哼,别人问的问题,看的我也头晕,百度了一下动态规划,看了看才想起来该怎么做,今天写了写代码,实现了~ 要求是递归,动态规划,想了想这种方法也是最简单的~ 所谓动态规划:把多阶段过程转化为一系列单阶段 ...

  10. c++ 函数调用在进入下一个循环的时候会再次初始化参数,将函数体直接写进去就正常

    #include"stdafx.h" #include"string" #include<iostream> #include<vector& ...