poj 1789 Truck History(最小生成树)

最新推荐文章于 2018-08-13 17:14:44 发布

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本文介绍了一种通过计算不同卡车类型代码之间的差异来确定最佳衍生计划的方法。利用Prim算法找到这些类型之间的最小生成树，从而得出最高可能的质量值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题意：有n个字符串，编号为1-n，问编号为2-n的字符串由总和的距离最小是多少，两两之间的距离为对应位置字母不同的个数，所以两两字符串之间的权值就是对应位置字母不同的个数，然后再最小生成树就行了，但稠密图用prim算法，稀疏图用kruskal算法，因为这个是稠密图，所以用prim算法。

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_{(t_o,t_d)}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.

AC代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<queue>
#include<set>
#define INF 999999999
#define LL long long
#define mod 1000003
#define MAX 30005
char a[2005][10];
int map[2005][2005];
int n;
int wei(int x, int y)
{
	int ans = 0;
	for (int i = 0; i < 7; i++)
	{
		if (a[x][i] != a[y][i])
			ans++;
	}
	return ans;
}
int prim()
{
	bool vis[2005];
	int lowdist[2005];
	for (int i = 1; i <= n; i++)
		lowdist[i] = 10, vis[i] = false;
	int s = 1;
	int m = 1;
	int ans = 0;
	int k;
	while (true)
	{
		if (m == n)
			break;
		int minn = 10;
		for (int j = 2; j <= n; j++)
		{
			if (!vis[j] && lowdist[j] > map[s][j])
				lowdist[j] = map[s][j];
			if (!vis[j] && minn > lowdist[j])
			{
				minn = lowdist[j];
				k = j;
			}
		}
		s = k;
		vis[k] = 1;
		ans += minn;
		m++;
	}
	return ans;
}
int main()
{
	while (scanf("%d", &n) && n)
	{
		for (int i = 1; i <= n; i++)
		{
			scanf("%s", a + i);
		}
		for (int i = 1; i < n; i++)
			for (int j = i + 1; j <= n; j++)
			{
				map[i][j] = map[j][i] = wei(i, j);
			}
		printf("The highest possible quality is 1/%d.\n", prim());
	}
}