POJ 1789 Truck History Prim算法模板题+处理下输入数据

我就是我 2024-02-19 18:30 42阅读 0赞

Truck History

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 2000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong> 65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 35112</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 13541</td> 
  </tr> 
 </tbody> 
</table>

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.

Source

[CTU Open 2003][]

题意解读：

货车的代码差值表示他们之间的距离，如aaaaaaa和aaaaaab，之间距离为1.

求连通这些距离的最小代价。模板题，难点：英文比较难理解，处理下输入数据。

我用的是最普通的方法处理输入数据的

so 代价：

![20190307141026167.png][]

#include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    #define range 2000
    #define inf 0x3f3f3f3f
    int map[range][range];
    bool vis[range];
    int dis[range];
    char s[range][7];
    int Dvalue(char a[],char b[])
    {
    	int i,j;
    	int sum=0;
    	for(i=0,j=0;i<7&&j<7;i++,j++)
    	{
    		if(a[i]!=b[j])
    		sum++;	
    	}
    	return sum;
    }
    int Prim(int n)
    {
    	int sum=0;
    	int pos=0;
    	int i,j;
    	vis[pos]=true;
    	for(i=0;i<n;i++)
    	{
    		dis[i]=map[pos][i];
    	}
    	for(i=0;i<n-1;i++)
    	{
    		int min=inf;
    		for(j=0;j<n;j++)
    		{
    			if(!vis[j]&&min>dis[j])
    			{
    				min=dis[j];
    				pos=j;
    			}
    		}
    		sum+=min;
    		vis[pos]=true;
    		for(j=0;j<n;j++)
    		{
    			if(!vis[j]&&dis[j]>map[pos][j])
    			dis[j]=map[pos][j];
    		}
    	}
    	return sum;
    }
    int main()
    {
    	int n;
    	while(~scanf("%d",&n),n)
    	{
    		for(int i=0;i<n;i++)
    		scanf("%s",s[i]);
    		memset(map,inf,sizeof(map));
    		memset(vis,false,sizeof(vis));
    		memset(dis,inf,sizeof(dis));
    		for(int i=0;i<n;i++)
    		{
    			for(int j=0;j<n;j++)
    			{
    				
    			
    					map[i][j]=map[j][i]=Dvalue(s[i],s[j]);
    			
    			}
    		}
    		
    		printf("The highest possible quality is 1/%d.\n",Prim(n));
    	}
    	return 0;
    }

看了看评论区，都是我这样处理数据的。

哪位大佬有效率更高的处理数据的方法，请评论区留思路。欢迎━(\*｀∀´\*)ノ亻!

[CTU Open 2003]: http://poj.org/searchproblem?field=source&key=CTU+Open+2003
[20190307141026167.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/01/29/870fb6d9bcb34595a859968916b565e9.png