poj1789 Truck History

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18828		Accepted: 7263

DescriptionAdvanced 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 to is the original type and td the type derived from it and d(to,td) 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. InputThe 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.OutputFor 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（借鉴自poj某神牛牪犇）： [code]题意大概是这样的：用一个7位的string代表一个编号，两个编号之间的distance代表这两个编号之间不同字母的个数。一个编号只能由另一个编号“衍生”出来，代价是这两个编号之间相应的distance，现在要找出一个“衍生”方案，使得总代价最小，也就是distance之和最小。例如有如下4个编号：aaaaaaabaaaaaaabaaaaaaabaaaa显然的，第二，第三和第四编号分别从第一编号衍生出来的代价最小，因为第二，第三和第四编号分别与第一编号只有一个字母是不同的，相应的distance都是1，加起来是3。也就是最小代价为3。

理解题意后思路很清晰，就是找最小生成树

先找出没两点间的权，然后按权由小到大排序即边的排序

然后就是按顺序找该边是否会构成回路，也就是并查集~

有一个剪纸，就是计算找到合适的边，等于n-1（即点数-1）的时候说明已经构成最小生成树

#include<iostream>#include<cstring>#include<algorithm>#include<string>using namespace std;int num,n,sum,t,arr[2000],ans;string s[2000];struct Node{int i,j,d;};Node node[2001005];bool cmp(const Node &a,const Node &b){return a.d<b.d;}int  _find(int x){return x==arr[x] ? x:arr[x]=_find(arr[x]); }int _unior(int x,int y){int xx=_find(x);int yy=_find(y);if(xx!=yy){ arr[xx]=yy;ans++; return 1;}else return 0;}void test(){  for(int i=0;i<t;i++){        if(_unior(node[i].i,node[i].j)){  sum+=node[i].d;  }      if(ans==n-1) break;  }}int main(void){while(cin>>n&&n){        ans=0;        sum=0; for(int i=0;i<n;i++){ cin>>s[i]; arr[i]=i;}   t=0;  for(int i=0;i<n;i++){    for(int j=i+1;j<n;j++){            num=0;        for(int k=0;k<7;k++){            if(s[i][k]!=s[j][k]) num++;        }        node[t].i=i;node[t].j=j;node[t].d=num;t++;    }  }   sort(node,node+t,cmp);   test();  cout<<"The highest possible quality is 1/"<<sum<<"."<<endl;}}