POJ 1789 Truck History

Truck History

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 26485		Accepted: 10279

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_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

题目大意时间：n个车牌号，刚开始只有一个车牌，其他车牌都是由一个车牌直接或间接产生 一个车牌到另一个车牌的产生权值是它们之间的数字不同的个数 问产生的最小的边权和

 1 #include<cstring>
 2 #include<iostream>
 3 #include<cstdio>
 4 using namespace std;
 5 const int maxn=2010;
 6 const int INF=0x3f;
 7 int map[maxn][maxn],dis[maxn],n;
 8 bool vis[maxn];
 9 char str[maxn][15];
10 int solve(){
11     for(int i=1;i<=n;i++) dis[i]=INF;
12     dis[1]=0;
13     memset(vis,0,sizeof(vis));
14     int MST=0;
15     while(1){
16         int Mix=INF,pos;
17         for(int i=1;i<=n;i++)
18           if(!vis[i]&&dis[i]<Mix)
19             Mix=dis[ pos=i ];
20         if(Mix == INF) break;
21         vis[pos]=true;
22         MST+=Mix;
23         for(int j=1;j<=n;j++)
24           dis[j]=min(dis[j],map[pos][j]);
25     }
26     return MST;
27 }
28 int main()
29 {
30     while(scanf("%d",&n)==1&& n!=0){
31         for(int i=1;i<=n;i++){
32             scanf("%s",str[i]+1);
33             for(int j=1;j<i;j++){
34                 int sum=0;
35                 for(int k=1;k<=7;k++)
36                   if(str[i][k] != str[j][k]) sum++;
37                 map[i][j]=map[j][i]=sum;
38             }
39         }
40         int ans=solve();
41         printf("The highest possible quality is 1/%d.\n" , ans);
42     }
43     return 0;
44 }

思路：见图，跑最小生成树，据说不加优化的Prim算法最优秀

Kursual版：

 

#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=2010;
const int INF=0x3f;
int n,tot,fa[maxn];
struct Edge{
    int from,to,value;
    bool operator < (const Edge &a) const{
        return value<a.value;
    }
}e[2000500];
char str[maxn][15];
int find(int x){
    if(x==fa[x]) return x;
    else return fa[x]=find(fa[x]);
}
int main()
{
    while(scanf("%d",&n)){
        if(n == 0) break;
        int Mst=0;tot=0;
        memset(e,0,sizeof(e));
        for(int i=1;i<=n;i++) scanf("%s",str[i]+1);
        for(int i=1;i<=n;i++) fa[i]=i;
        for(int i=1;i<n;i++){
            for(int j=i+1;j<=n;j++){
                int sum=0;
                for(int k=1;k<=7;k++)
                  if(str[i][k] != str[j][k]) sum++;
                e[++tot].from=i;e[tot].to=j;e[tot].value=sum;
            }
        }
        sort(e+1,e+tot+1);
        int cur=0;
        for(int i=1;i<=tot;i++){
            int u=e[i].from,v=e[i].to,w=e[i].value;
            int rx=find(u),ry=find(v);
            if(rx!=ry){
                fa[rx]=ry;
                Mst+=w;cur++;
            }
            if(cur == n-1) break;
        }
        printf("The highest possible quality is 1/%d.\n" ,Mst);
    }
    return 0;
}

 

注：上面标红色的那一句，我刚开始写在定义循环变量j那里，会导致WA掉。。。