Truck History---poj1789

题目链接：http://poj.org/problem?id=1789

 

题意： They defined the distance of truck types as the number of positions with different letters in truck type codes

 

这句话说明距离被定义成编号之间字母的不同个数；编号是由7个字符构成的字符串；

 

然后求最小生成树；

 

#include<stdio.h>
#include<string.h>
#include<map>
#include<iostream>
#include<algorithm>
#include<math.h>
#define N 2010
#define INF 0xfffffff

using namespace std;

int maps[N][N],vis[N],dist[N], n;

int Prim(int start)
{
    int ans=0;
    vis[start] = 1;
    for(int i=1; i<=n; i++)
    {
        dist[i] = maps[start][i];
    }
    for(int i=1; i<=n; i++)
    {
        int Min = INF, index = -1;
        for(int j = 1; j <= n; j++)
        {
            if(vis[j]==0 && Min>dist[j])
            {
                Min = dist[j];
                index = j;
            }
        }
        if(index == -1) break;
        ans += Min;
        vis[index] = 1;
        for(int j=1; j<=n; j++)
        {
            if(vis[j]==0 && dist[j] > maps[index][j])
                dist[j] = maps[index][j];
        }
    }
    return ans;
}

int main()
{
    int i, j, k;
    char s[N][20];
    while(scanf("%d", &n), n)
    {
        for(i=1; i<=n; i++)
        {
            scanf("%s", s[i]);
            vis[i] = 0;
            dist[i] = INF;
            for(j=1; j<=n; j++)
                maps[i][j] = INF;
        }

        for(i=1; i<=n; i++)
        {
            for(j=1; j<i; j++)
            {
                int cnt = 0;
                for(k=0; k<7; k++)
                    if(s[i][k] != s[j][k])
                        cnt++;
                maps[i][j] = maps[j][i] = cnt;
            }
        }
        int ans;
        ans = Prim(1);
        printf("The highest possible quality is 1/%d.\n",  ans);
    }
    return 0;
}