Truck History （Prim算法）

Truck History

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 79   Accepted Submission(s) : 27

Problem 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(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 <= 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

4aaaaaaabaaaaaaabaaaaaaabaaaa0

 

Sample Output

The highest possible quality is 1/3.

 

Source

PKU

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：

用prim算法，一开始套的模板结果超市，后来修改了一点，Ac了。。。。

代码：

#include <iostream>#include <cstring>#include <string>#include <stdio.h>#define inf 0x3f3f3f3fusing namespace std;int g[3000][3000];int minn[3000];bool u[3000];string ss[3000];int main(){    //ios::sync_with_stdio(false);    int n,i,j;    while(scanf("%d",&n),n){        for(i=1;i<=n;i++)        {            cin>>ss[i];            for(j=1;j<=i;j++)            {                if(i==j){g[i][j]=0;continue;}                int tt=0;                for(int k=0;k<7;k++){if(ss[i][k]!=ss[j][k])tt++;}//记录距离                g[i][j]=g[j][i]=tt;            }        }        memset(minn,0x7f,sizeof(minn));        minn[1]=0;        memset(u,1,sizeof(u));        //int sum=0;        int cnt=inf;        for(i=1;i<=n;i++)        {            int k=0;            for(j=1;j<=n;j++)//找一个与生成树点相连的不在生成树的最小的点                if(u[j]&&(minn[j]<minn[k]))k=j;            u[k]=0;  //k入树            for(j=1;j<=n;j++) //修改所有与k相连的未进入生成树的点                if(u[j]&&(g[k][j]<minn[j]))                minn[j]=g[k][j];        }        int sum=0;        for(i=1;i<=n;i++)sum+=minn[i]; //累加权值        printf("The highest possible quality is 1/%d.\n",sum);    }return 0;}