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

来源：互联网发布：曼哈顿软件招聘编辑：程序博客网时间：2024/05/21 19:13

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);   // MSTint sst = INF + 1;    //SST 初始化for(int i = 1; i <= M; i++){ //求SSTif(map[i].mark == 1){  //这条边属于MSTmst(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MSTint 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 MSTint 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;}}}

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