最小生成树kruskal算法

#include<cstdio>
#include<algorithm>
#include <iostream>
using namespace std;

const int maxn = 2000+10;
char map[maxn][10];
int f[maxn];
int n,cnt;

struct edge{
    int u,v;
    int w;
}e[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 == f[x] ? x : f[x] = find(f[x]);
}


int Kruskal()
{
    int ans = 0;
    for(int i = 1; i <= n; i++) f[i] = i;
    sort(e,e+cnt,cmp);

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

        if(u != v)
        {
            f[v] = u;
            ans += e[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]);

        cnt = 0;
        for(int i = 1; i < n; i++)
        {
            for(int j = i+1; j <= n; j++)
            {
                e[cnt].u = i;
                e[cnt].v = j;
                e[cnt].w = dist(i,j); //把字符串差值变权值
                cnt++;
            }
        }

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

dist为处理权值的函数，也可以把判断的连通问题写进merge函数，并查集压缩路径，用来提高效率