POJ 1789 Truck History (Prim算法求最小生成树)

题目

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
$$\frac{1}{\sum_{(t_o,t_d)}d(t_o,t_d)}$$

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, $N$ , $2 \leq N \leq 2000$. 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 $\frac{1}{Q}$ .", where $\frac{1}{Q}$ 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.

题意

历史上,曾用 $7$ 个小写字母来表示每种truck的型号,每两种型号之间的差距为字母串中不同字母的个数。现在给出 $n$ 种不同型号的truck,问怎样使
$\frac{1}{\sum_{(t_o,t_d)}d(t_o,t_d)}$ 的值最小。(即找到一条连接所有truck的最短路径。典型的最小生成树的问题,$Prim$ 算法适合稠密图,$Kruskal$ 算法适合稀疏图,可以使用 $Prim$ 和 $kruskal$ 两种方法。该题是稠密的图。


分析

输入的字符串需要转换成邻接矩阵,然后直接套 $Prim$ 模板就好啦~


代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<cmath>
#include<set>
#include<cstdlib>
#include<functional>
#include<climits>
#include<cctype>
#include<iomanip>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define mod 1e9+7
#define clr(a,x) memset(a,x,sizeof(a))
const double eps = 1e-6;
const int MAXN=2010;
bool vis[MAXN];
int lowc[MAXN];
int map[MAXN][MAXN];
int Prim(int cost[][MAXN],int n)
{
    int ans=0;
    clr(vis,0);
    vis[0]=1;
    for(int i=1;i<n;i++)
    {
        lowc[i]=cost[0][i];
    }
    for(int i=1;i<n;i++)
    {
        int minc=INF;
        int p=-1;
        for(int j=0;j<n;j++)
            if(!vis[j]&&minc>lowc[j])
            {
                minc=lowc[j];
                p=j;
            }
        if(minc==INF)
        {
            return -1;
        }
        vis[p]=1;
        ans+=minc;
        for(int j=0;j<n;j++)
            if(!vis[j]&&lowc[j]>cost[p][j])
                lowc[j]=cost[p][j];
    }
    return ans;
}
int main()
{
    int t;
    char str[MAXN][8];
    int edge;
    while(cin>>t,t)
    {
        for(int i=0;i<t;i++)
        {
            scanf("%s",&str[i]);
        }
        for(int i=0;i<t;i++)  //接下来就是建立邻接矩阵
        {
            for(int j=0;j<t;j++)  //这里wa了好久
            {
                edge=0;
                for(int k=0;k<7;k++)
                {
                    if(str[i][k]!=str[j][k])
                        edge++;
                    map[i][j]=map[j][i]=edge;
                }
            }
        }
        int ans;
        ans=Prim(map,t);  //Prim算法
        printf("The highest possible quality is 1/%d.\n",ans);
    }
    return 0;
}

PS:太菜,太菜了哇...

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Last modification:November 1st, 2019 at 12:57 am
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