Truck History(最小生成树_prim算法)

Truck HistoryCrawling in process...Crawling failedTimeLimit:2000MS Memory Limit:65536KB64bit IO Format:%I64d & %I64uSubmitStatusPracticePOJ 1789DescriptionAdvanced 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 stringof 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 fromthe 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 differentletters 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 as1/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.InputThe 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 lowercaseletters). 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.OutputFor 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：给出n个长度为7的字符串，每个字符串代表一个车，定义车的距离是两个字符串间不同字母的个数，题目要求的数不同的车的距离的最小值，即所求的就是最小生成树

思路：裸的普林母算法。只是注意开始先求出距离（即相同位置的不同字母的个数）

#include <stdio.h>#include <string.h>#include <stdlib.h>#define inf 99999999int map[2010][2010];char str[2010][10];int dis[2010];//记录边的权值int flag[2010];//记录是否被访问int find(int a,int b){int i;int count=0;for(i=0;i<7;i++){if(str[a][i]!=str[b][i]){count++;}}return count;}int prim(int n){int max,i,j,k,sum=0;memset(flag,0,sizeof(flag));for(i=0;i<n;i++)dis[i]=map[0][i];flag[0]=1;//零点标记为已经访问for(i=1;i<n;i++)//进行n-1次操作{max=inf;//初始化for(j=1;j<n;j++)//遍历所有点if(max>dis[j]&&!flag[j])//在所有未加入的点中找一个最小的权值{k=j;//记录下标max=dis[j];//更新最小值}if(max==inf)//若图是不连通的break;//提前退出flag[k]=1;sum+=max;for(j=1;j<n;j++)//遍历所有的点if(dis[j]>map[k][j]&&!flag[j])//对未加入的点&&能找到与此点相连且的权值最小的边dis[j]=map[k][j];//进行更新}printf("The highest possible quality is 1/%d.\n",sum);}int main(){int n,i,j;while (~scanf("%d",&n)){memset(map,0,sizeof(map));memset(str,0,sizeof(str));if(n==0)break;for(i=0;i<n;i++)scanf("%s",str[i]);for(i=0;i<n;i++)for(j=0;j<i;j++)map[i][j]=map[j][i]=find(i,j);prim(n);}return 0;}