UVA 10600 ACM Contest and Blackout(次小生成树)

题意：

　　给你点和边，求出最小生成树 和  次小生成树。

思路：

　　先求一次最小生成树，然后标记每一条边，依次删除，再求最小生成树，从中找到最小的就是次小生成树了。

代码：   

import java.util.Scanner;
import java.util.Comparator;
import java.util.Arrays;

class Node{
    public int u, v, w, mark;
}
//结构排序
class mycmp implements  Comparator<Node>{
    public int compare(Node A, Node B){ 
                return A.w - B.w;  
      }  
}
public class Main {
    final static int MAXN = 10000 + 13;
    final static int INF = 0x3f3f3f3f;
    static int[] pre = new int[MAXN];
    static Node[] map = new Node[MAXN];
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while(T != 0){
            int N,M;
            N = sc.nextInt();
            M = sc.nextInt();
            for(int i = 1; i <= M; i++){
                map[i]=new Node();  
                map[i].u = sc.nextInt();
                map[i].v = sc.nextInt();
                map[i].w = sc.nextInt();
                map[i].mark = 0;
            }
            mst(N);
            Arrays.sort(map, 1, M + 1, new mycmp());
            int mst = ksu(N, M);   // MST
            int sst = INF + 1;    //SST 初始化
            for(int i = 1; i <= M; i++){ //求SST
                if(map[i].mark == 1){  //这条边属于MST
                    mst(N);
                    int temp = ksu(N, M, i);       //删除一条边后得到的结果、如果大于 0 说明构造成功，否则构造失败
                    if(temp < sst && temp != -1){ 
                        sst = temp;
                    }
                }
            }
            System.out.println(mst + " " + sst);
            T--;
        }
        sc.close();
    }
    public static int ksu(int N, int M){ //求MST
        int cnt = 0;
        int ans= 0;
        for(int i = 1; i <= M; i++){
            int fu = Find(map[i].u);
            int fv = Find(map[i].v);
            if(fu != fv){
                ans += map[i].w;
                cnt++;
                pre[fv] = fu;
                map[i].mark = 1;   //标记
            }
            if(cnt == N - 1){
                return ans;
            }
        }
        return ans;
    }
    public static int ksu(int N, int M,int mark){  //删除 mark 这条边 求 MST
        int ans= 0;
        int cnt = 0;
        for(int i = 1; i <= M; i++){
            if(i == mark) continue;    //删除
            int fu = Find(map[i].u);
            int fv = Find(map[i].v);
            if(fu != fv){
                ans += map[i].w;
                cnt++;
                pre[fv] = fu;
            }
            if(cnt == N - 1){
                return ans;
            }
        }
        return -1;
    }
    public static int Find(int x){
        return x == pre[x] ? x : (pre[x] = Find(pre[x]));
    }
    public static void debug(int M){
        for(int i = 1; i <= M; i++){
            System.out.println(i + " " + map[i].u + " " + map[i].v + " " + map [i].w + " "+ map[i].mark);
        }
    }
    public static void mst(int N){
        for(int i = 1; i <= N; i++){
            pre[i] = i;
        }
    }
}