POJ - 1789 Truck History （最小生成树）

青旅半醒 2022-05-18 06:57 171阅读 0赞

#                                             Truck History #

# 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.

**题意描述：**  
每个字符串有7个字符组成，除了第一个其他有别的字符串衍生出得，权值为字符串中同位置不同字符的数量，求最小生成树。

**解题思路：**

求出每两个字符串间的权值，存入map数组，用prim算法求出最小生成树。

#include<stdio.h>
    #include<string.h>
    # define inf 0x7f7f7f7f
    int map[3010][3010],book[3010],dis[3010];
    char a[3010][10];
    int main()
    {
    	int n,min,i,j,k,u,v,num,sum;
    	while(scanf("%d\n",&n))
    	{
    		if(n==0)
    			break;
    		for(i=0;i<n;i++)
    			gets(a[i]);
    		memset(map,0,sizeof(map));
    		//求两个字符串之间的权值，存入map数组 
    		for(i=0;i<n-1;i++)
    			for(j=i+1;j<n;j++)
    			{
    				num=0;
    				for(k=0;k<7;k++)
    				{
    					if(a[i][k]!=a[j][k])
    						num++;
    				}
    				map[i+1][j+1]=num;
    				map[j+1][i+1]=num; 
    			}
    		for(i=1;i<=n;i++)
    			dis[i]=map[1][i];
    		memset(book,0,sizeof(book));
    		book[1]=1;
    		sum=0;
    		//prim
    		for(i=1;i<n;i++)
    		{
    			min=inf;
    			for(j=1;j<=n;j++)
    			{
    				if(book[j]==0&&dis[j]<min)
    				{
    					min=dis[j];
    					u=j;
    				}
    			}
    			book[u]=1;
    			sum=sum+dis[u];
    			for(v=1;v<=n;v++)
    			{
    				if(book[v]==0&&dis[v]>map[u][v])
    					dis[v]=map[u][v];
    			}
    		}
    		printf("The highest possible quality is 1/%d.\n",sum);
    	}
    	return 0;
    }