POJ 1789 Truck History 最小生成树 KRUSKAL算法

这题的题意并不难，就是根据不同车型的代号分别求出两个代号的“距离”

然后根据距离选择一个最小的生成树，下面用Kruskal算法实现，虽然AC了，资源占用很大，还需改进

Truck History

Time Limit:2000MS		Memory Limit:65536K

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_ois the original type and t_dthe 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.

下面使用Kruskal算法

/* Author yan
 * POJ 1789
 * Truck History
*/
#include<stdio.h>
#define MAX 2000
int n;
char type[MAX][8];
typedef struct _route
{
    int x,y;
    int weight;
};
struct _route route[1999000];

int sum;

int father[MAX];
int rank[MAX];

void initial()
{
	int _i;
	memset(rank,0,sizeof(rank));
	for(_i=0;_i<n;_i++)
		father[_i]=_i;
}
int Find_Set(int x)
{
	if(x!=father[x]) father[x]=Find_Set(father[x]);
	return father[x];
}
int Union_Set(int x,int y,int w)
{
	int a,b;
    a=Find_Set(x);
    b=Find_Set(y);
    if(a==b) return 0;
    if(rank[x]>rank[y])
        father[b]=a;
    else
    {
        father[a]=b;
        if(rank[x]==rank[y]) rank[x]++;
    }
    sum+=w;
    return 1;
}
/*
下面这种参数传递方法导致TLE超时,修改后AC
int cmp(struct _route a,struct _route b)
{
	return a.weight - b.weight;
}
*/
int cmp(const void * a,const void * b)
{
	return (*(struct _route *)a).weight - (*(struct _route *)b).weight;
}

int main()
{
	int i,j,k;
	int cnt,count;
	int x,y;
	freopen("input","r",stdin);
	while(scanf("%d",&n)&&n!=0)
	{
		count=0;
		sum=0;
		initial();
		for(i=0;i<n;i++) scanf("%s",type[i]);
		for(i=0;i<n;i++)
			for(j=i+1;j<n;j++)
			{
				cnt=0;
				for(k=0;k<7;k++)
				{
					if(type[i][k]!=type[j][k]) cnt++;
				}
				route[count].x=i,route[count].y=j,route[count].weight=cnt;
				//printf("%d %d %d/n",i,j,cnt);
				count++;
			}
		qsort(route,count,sizeof(route[0]),cmp);
		for(i=0;i<count;i++)
		{
/*
			x=Find_Set(route[i].x);
			y=Find_Set(route[i].y);
			//printf("x=%d  y=%d/n",route[i].x,route[i].y);
			if(x!=y) Union_Set(x,y,route[i].weight);
*/
			Union_Set(x,y,route[i].weight);
		}
		printf("The highest possible quality is 1/%d./n",sum);
	}
	return 0;
}