算法分析

普通的最小生成树：所有边权之和最小

本题的最小生成树：最大的边权最小

在这里我深刻的意识到kruskal本质是让点按照某个规律进行连通，res是特有的属性

做法：kruskal

• 1、将所有边从小到大排序

• 2、枚举每条边a,b,权值是w

  • if a和b不连通，将a，b加进集合，且更新res

时间复杂度 $O(mlogm)$

参考文献

算法提高课

Java 代码

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {
    static int N = 310, M = 10010;
    static int n,m;
    static Edge[] edge = new Edge[M];
    static int[] p = new int[N];
    static int find(int x)
    {
        if(p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
    static int kruskal()
    {
        int res = 0;
        Arrays.sort(edge,0,m);

        for(int i = 0;i < m;i ++)
        {
            int a = edge[i].a;
            int b = edge[i].b;
            int w = edge[i].w;
            a = find(a);
            b = find(b);
            if(a != b)
            {
                p[a] = b;
                res = w;
            }
        }
        return res;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] s1 = br.readLine().split(" ");
        n = Integer.parseInt(s1[0]);
        m = Integer.parseInt(s1[1]);
        for(int i = 0;i < m;i ++)
        {
            String[] s2 = br.readLine().split(" ");
            int a = Integer.parseInt(s2[0]);
            int b = Integer.parseInt(s2[1]);
            int w = Integer.parseInt(s2[2]);
            edge[i] = new Edge(a,b,w);
        }

        for(int i = 1;i <= n;i ++) p[i] = i;

        System.out.println(n - 1 + " " + kruskal());

    }
}
class Edge implements Comparable<Edge>
{
    int a,b,w;
    Edge(int a,int b,int w)
    {
        this.a = a;
        this.b = b;
        this.w = w;
    }
    @Override
    public int compareTo(Edge o) {
        // TODO 自动生成的方法存根
        return Integer.compare(w, o.w);
    }
}