POJ 1789 Truck History[최소 생 성 트 리 단순 응용]

링크:
http://poj.org/problem?id=1789
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/G
Truck History
Time Limit: 2000MS
 
Memory Limit: 65536K
Total Submissions: 14950
 
Accepted: 5714
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(to,td)
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
제목:
             
  최소한 의 거리 와
        문자열 마다 7 글자 만 있 습 니 다.
        두 문자열 의 거 리 는 대응 하 는 위치 가 다른 문자 개 수 를 세 는 것 이다.
         
알고리즘:최소 생 성 트 리
생각:
               
 모든 문자열 을 하나의 장소 로 보고 문자열 간 의 서로 다른 문자 개 수 를 장소 간 의 거리 로 본다.최소 생 성 트 리 를 사용 하면 됩 니 다.
Kruskal：
1789
Accepted
22860K
563MS
C++
1378B

//Accepted	22860 KB	579 ms	C++	1302 B	2013-07-31 09:37:35
#include<stdio.h>
#include<algorithm>
using namespace std;

const int maxn = 2000+10;
char map[maxn][10];
int p[maxn];
int n,m;

struct Edge{
    int u,v;
    int w;
}edge[maxn*maxn/2];

int dist(int st, int en)
{
    int distance = 0;
    for(int i = 0; i < 7; i++)
        if(map[st][i] != map[en][i])
            distance++;
    return distance;
}

bool cmp(Edge a, Edge b)
{
    return a.w < b.w;
}

int find(int x)
{
    return x == p[x] ? x : p[x] = find(p[x]);
}

int Kruskal()
{
    int ans = 0;
    for(int i = 1; i <= n; i++) p[i] = i;
    sort(edge,edge+m,cmp);

    for(int i = 0; i < m; i++)
    {
        int u = find(edge[i].u);
        int v = find(edge[i].v);

        if(u != v)
        {
            p[v] = u;
            ans += edge[i].w;
        }
    }
    return ans;
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) break;
        for(int i = 1; i <= n; i++)
            scanf("%s", map[i]);

        m = 0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = i+1; j <= n; j++)
            {
                edge[m].u = i;
                edge[m].v = j;
                edge[m++].w = dist(i,j);
            }
        }

        int ans = Kruskal();
        printf("The highest possible quality is 1/%d.
", ans);
    }
    return 0;
}

Prime：
Accepted
15672K
454MS
C++
1289B
2013-07-31 09:38:56

//Accepted	15672 KB	469 ms	C++	1227 B	2013-07-31 09:37:25
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn = 2000+10;
const int INF = maxn*7;

char map[maxn][10];
int w[maxn][maxn];
int d[maxn];
int vis[maxn];
int n;

int dist(int st, int en)
{
    int distance = 0;
    for(int i = 0; i < 7; i++)
        if(map[st][i] != map[en][i])
            distance++;
    return distance;
}

int Prime()
{
    int ans = 0;
    for(int i = 1; i <= n; i++) d[i] = INF;
    d[1] = 0;
    memset(vis, 0, sizeof(vis));

    for(int i = 1; i <= n; i++)
    {
        int x, m = INF;
        for(int y = 1; y <= n; y++) if(!vis[y] && d[y] <= m) m = d[x=y];
        vis[x] = 1; ans += d[x];
        for(int y = 1; y <= n; y++) if(!vis[y])
            d[y] = min(d[y], w[x][y]);
    }
    return ans;
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) break;

        for(int i = 1; i <= n; i++) scanf("%s", map[i]);

        for(int i = 1; i <= n; i++)
        {
             for(int j = i+1; j <= n; j++)
             {
                 w[i][j] = dist(i,j);
                 w[j][i] = w[i][j];
             }
        }

        int ans = Prime();
        printf("The highest possible quality is 1/%d.
", ans);
    }
    return 0;
}