poj 1789 Truck History

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最新推荐文章于 2023-01-10 18:14:39 发布

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分类专栏：最小生成树文章标签：最小生成树

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解决一个特定的历史问题，即如何确定一组卡车类型的最优衍生路径。通过计算不同卡车类型间的距离（基于其代码中不同的字符数量），并采用最小生成树算法来找到使衍生计划质量最高的方案。

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题目：

Truck History

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1789

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

aaaaaaa

baaaaaa

abaaaaa

aabaaaa

题目大意：

给n个7位的字符串,求最小的距离和,距离为不相同字符的个数

题目思路：

1、最小生成树的思路

程序：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cctype>
#include <fstream>
#include <limits>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cassert>
using namespace std;
bool v[4000];
char s[2010][10];
struct node
{
    double z;
    int w;
    friend bool operator < (node b,node c)
    {
        return b.z>c.z;
    }
}b,c;
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        memset(v,0,sizeof(v));
        priority_queue<node>qu;
        for(int i=0;i<n;i++)
            scanf("%s",&s[i]);
        b.w=0;b.z=0;
        qu.push(b);
        int time =0,ans=0;
        while(~qu.empty())
        {
            b=qu.top();
            qu.pop();
            if(v[b.w]) continue;
            v[b.w]=1;
            ans+=b.z;
            time++;
            if(time==n)
                break;
            for(int i=0;i<n;i++)
                if(!v[i])
                {
                    c.w=i;
                    c.z=0;
                    for(int j=0;j<7;j++)
                        if(s[i][j]!=s[b.w][j])
                        c.z++;
                    qu.push(c);
                }
        }
        printf("The highest possible quality is 1/%d.\n",ans);
    }
return 0;
}