prime（最小生成树）—

prime（最小生成树）——POJ 1789

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Truck History

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1789

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_{(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.

题意：就是给你一些车牌号，然后你通过比较两个车牌号相同位置字符不同的个数作为权值，再一个完全图的最小生成树算法。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=2000+10;
const int INF=1<<30;
using namespace std;
int G[MAXN][MAXN];
int cp[MAXN];
int dis[MAXN];
int sum;
char truck[MAXN][10];void prim(int n, int v)
{int i,j;for(i=0; i<n; i++){dis[i]=G[v][i];cp[i]=v;}for(i=1; i<n; i++){int min=INF;int k=v;for(j=0; j<n; j++){if(dis[j] && dis[j]<min){min=dis[j];k=j;}}//printf("%d <-> %d = %d\n", cp[k], k, min);//打印两顶点与两点间权值sum+=dis[k];//计算最短路径和dis[k]=0;//加入集合for(j=0; j<n; j++){if(G[j][k] && G[j][k]<dis[j]){//更新候选边dis[j]=G[j][k];cp[j]=k;}}}printf("The highest possible quality is 1/%d.\n", sum);
}int Distance(char const *str1, char const *str2)
{const char *p;const char *q;int cnt=0;for(p = str1, q = str2; *p && *q;)if(*p++ != *q++) cnt++;return cnt;
}int main()
{//freopen("in.txt","r",stdin);int n;while(scanf("%d", &n), n)//节点为0~n-1{sum=0;int i,j;for(i=0; i<n; i++) scanf("%s", truck[i]);for(i=0; i<n; i++){//初始化图for(j=0; j<n; j++){if(i==j){ G[i][j]=0; continue;}G[i][j]=INF;}}for(i=0; i<n; i++){for(j=i+1; j<n; j++){int len = Distance(truck[i], truck[j]);G[i][j] = G[j][i] = len;}}prim(n, 0);}return 0;
}

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