Truck History(prim)

http://poj.org/problem?id=1789

读不懂题再简单也不会做，英语是硬伤到哪都是真理，sad++。

此题就是一个最小生成树，两点之间的权值是毎两串之间的不同字母数。

#include<stdio.h>
#include<string.h>
const int N=2020;
const int INF=1<<28;
int map[N][N],vis[N],dis[N];
int n,sum;
void prim()
{
    int pos;
    for (int i = 1; i <= n; i ++)
    {
        dis[i] = map[1][i];
    }
    vis[1] = 1;
    for (int i = 1; i <= n-1; i ++)
    {
        int min = INF;
        for (int j = 1; j <= n; j ++)
        {
            if (!vis[j] && dis[j] < min)
            {
                min = dis[j];
                pos = j;
            }
        }
        sum += min;
        vis[pos] = 1;
        for (int j = 1; j <= n; j ++)
        {
            if (!vis[j] && dis[j] > map[pos][j])
                dis[j] = map[pos][j];
        }

    }
}
void init()
{
    sum = 0;
    for (int i = 0; i <= n; i ++)
    {
        for (int j = 0; j <= n; j ++)
        {
            map[i][j]  = INF;
        }
        map[i][i] = 0;
        vis[i] = 0;
    }

}
int main()
{
    char str[N][12];
    int cnt;
    while(~scanf("%d",&n)&&n)
    {

        init();
        for (int i = 1; i <= n; i ++)
        {
            scanf("%s",str[i]+1);
        }
        for (int i = 1; i <= n; i ++)
        {
            for (int j = i+1; j <= n; j ++)
            {
                cnt = 0;
                for (int k = 1; k <= 7; k ++)
                {
                    if (str[i][k]!=str[j][k])
                        cnt++;
                }
                map[i][j] = cnt;
                map[j][i] = cnt;
            }
        }
        prim();
        printf("The highest possible quality is 1/%d.
",sum);
    }
    return 0;

}