POJ1789 Truck History(Prim)

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24724		Accepted: 9636

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

4
aaaaaaa
baaaaaa
abaaaaa
aabaaaa
0

Sample Output

The highest possible quality is 1/3.

Source

CTU Open 2003

【题意】有N辆车，每辆车都有不同的字符串编码，而且都是7位。将编码的距离定义为编码字符串中（7个位置上）不同字符的位置数目。除了第一辆卡车，其他的卡车都是由另一类卡车派生出来的，再定义派生方案的优劣值为总距离值的倒数，问你这个倒数最大是多少。

【分析】倒数最大，即总距离值最小，很明显就是最小生成树，可用Prim。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=2005;
const int M=15005;
int edg[N][N];
int lowcost[N];//记录未加入树集合的i离树集合中元素最小的距离
char w[N][8];
int n,m,t;
int fun(char  *a,char *b)
{
    int cnt=0;
    for(int k=0;k<7;k++){
        if(a[k]!=b[k])cnt++;
    }return cnt;
}
void Build()
{
    for(int i=0;i<n;i++)
    {
        for(int j=i+1;j<n;j++)
        {
            edg[i][j]=edg[j][i]=fun(w[i],w[j]);
        }
    }
}
void prim()
{
    int sum=0;lowcost[0]=-1;
    for(int i=1;i<n;i++){
        lowcost[i]=edg[0][i];
    }
    for(int i=1;i<n;i++){
        int minn=inf,k;
        for(int j=0;j<n;j++){
            if(lowcost[j]!=-1&&lowcost[j]<minn){
                k=j;minn=lowcost[j];
            }
        }
        sum+=minn;
        lowcost[k]=-1;
        for(int j=0;j<n;j++){
            lowcost[j]=min(lowcost[j],edg[k][j]);
        }
    }
    printf("The highest possible quality is 1/%d.
",sum);
}
int main() {
    while(~scanf("%d",&n)&&n) {
        memset(edg,0,sizeof(edg));
        memset(lowcost,0,sizeof(lowcost));
        for(int i=0;i<n;i++){
            scanf("%s",w[i]);
        }
        Build();
        prim();
    }
    return 0;
}