POJ 1789 Truck History 最小生成树

Truck History Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 23796   Accepted: 9227 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 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

点击打开题目链接

n辆卡车，起初只有一辆，其他n-1辆是由原有的派生而来。

每辆卡车由一个长度为7的小写字母串组成，派生时的花费为两个字符串不同字符的位置个数。

最小生成树变形，Kruskal 算法。

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;

const int MAXN = 2100;
const int MAXM = 2100000;

char ch[MAXN][10];
int f[MAXN];

struct Edge
{
    int u, v, w;
    bool operator < (const Edge& t) const   //按质量从小到大排序
    {
        return w < t.w;
    }
} edge[MAXM];

int Find(int x)         //并查集，找到 x 的祖先
{
    return x == f[x] ? x : f[x] = Find(f[x]);
}

int main()
{
    int n;
    while (~scanf("%d", &n), n)
    {
        int sumOfEdge = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%s", ch[i]);
            for (int j = 0; j < i; j++) //保存所有的边
            {
                int len = 0;
                for (int k = 0; k < 7; k++) if (ch[i][k] != ch[j][k]) len++;
                edge[sumOfEdge].u = i;
                edge[sumOfEdge].v = j;
                edge[sumOfEdge++].w = len;
            }
        }
        sort(edge, edge + sumOfEdge);   //排序
        int ans = 0;                    //初始化MST为空
        for (int i = 0; i <= n; i++) f[i] = i;  //初始化连通分量，每个点自成一个独立的连通分量
        for (int i = 0; i < sumOfEdge; i++)
        {
            int x = Find(edge[i].u), y = Find(edge[i].v);
            if (x != y)         //u，v不在同一个连通分量
            {
                ans += edge[i].w;   //将edge[i]加入MST
                f[x] = y;           //合并u，v所在的连通分量
            }
        }
        printf("The highest possible quality is 1/%d.\n", ans);
    }
    return 0;
}