ZOJ2158，POJ1789

题意：给你n辆车，每辆车有7个字符的编号，他们之间的距离值等于他们中不同字符的位置数目，然后叫你求最高优劣值的派生方案 1/Q，Q就要最小，求最小生成树

       

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
#define inf 999999
#define N 2013
int n,dis[N],vis[N],g[N][N];
char s[N][10];

int prim(int st)
{
    for(int i=1; i<=n; i++)
    {
        dis[i] = g[i][st];
        vis[i] = 0;
    }
    dis[st] = 0 ; vis[st] = 1;
    int cost = 0;
    for(int T=1; T<n; T++)
    {
        int mindis = inf ,idx = -1;
        for(int i=1; i<=n; i++)
        {
            if(!vis[i] && dis[i] < mindis)
            {
                mindis = dis[i];
                idx = i;
            }
        }
        cost += mindis;
        if(idx==-1) return -1;
        vis[idx] = 1;
        for(int i=1; i<=n; i++)
        {
            if(!vis[i] && dis[i] > g[i][idx])
            dis[i] = g[i][idx];
        }
    }
    return cost;
}

int main()
{
    while(scanf("%d",&n)&&n)
    {
        for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
        {
            if(i==j) g[i][j] = 0;
            else g[i][j] = inf;
        }
        for(int i=1; i<=n; i++)
        cin>>s[i];
        for(int i=1; i<=n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                int cnt = 0;
                for(int k=0; k<7; k++)
                {
                    if(s[i][k]!=s[j][k])
                    cnt++;
                }
                g[i][j] = g[j][i] = cnt;
            }
        }
        printf("The highest possible quality is 1/%d.
",prim(1));
    }
    return 0;
}