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@ -1,102 +0,0 @@
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import java.util.Scanner;
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import java.util.Set;
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import java.util.TreeSet;
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将城镇抽象为顶点，将有道路联通的城镇划分到同一个集合中，将问题转化成求集合个数的问题
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public class ChangTongEngine {
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in);
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while (scanner.hasNextInt()) {
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int n = scanner.nextInt(); n表示城镇数量
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if (n == 0) {
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break;
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}
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int m = scanner.nextInt(); m表示道路数目
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初始化一个并查集
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UnionFind uf = new UnionFind(n+1);
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for (int i = 0; i m; i++) {
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m1和m2表示两个城镇编号
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int m1 = scanner.nextInt();
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int m2 = scanner.nextInt();
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合并已经建成的城市之间的道路
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uf.union(m1, m2);
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}
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TreeSet中元素是有序且不重复的
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TreeSetInteger ts = new TreeSetInteger();
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for (int i = 1; i = n; i++) {
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ts.add(uf.find(i));
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}
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n个节点至少需要n-1条边来连接
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System.out.println(ts.size() - 1);
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}
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}
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}
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并查集类
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class UnionFind{
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int[] parent; 记录结点的父亲
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int[] height; 记录树的高度
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有参数构造方法
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初始化并查集
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public UnionFind(int n) {
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parent = new int[n];
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height = new int[n];
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for (int i = 0; i parent.length; i++) {
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parent[i] = i; 初始时，每个节点的父亲节点都是自己
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height[i] = 0; 初始时，所有树高均为1
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}
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}
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查操作，找到元素x的根结点
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public int find(int x) {
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if (x != parent[x]) {
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压缩路径：先找到根结点，再将查找路径上所有结点都挂到根结点下
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parent[x] = find(parent[x]);
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}
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return parent[x];
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}
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合并操作
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public void union(int x,int y) {
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int rootX = find(x);
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int rootY = find(y);
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如果两颗树的根结点相同时，说明两个节点在一个集合中
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if (rootX == rootY) {
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return;
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}else {
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按高度合并
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if (height[rootX] height[rootY]) {
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parent[rootY] = rootX;
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}else if (height[rootX] height[rootY]) {
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parent[rootX] = rootY;
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}else{
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parent[rootY] = rootX;
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如果高度相同则合并后高度加1
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height[rootX]++;
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}
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}
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for (int i = 0; i parent.length; i++) {
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if (parent[i] == rootX) {
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遍历并查集中的结点，如果并查集中结点的祖先是x的祖先，则全部设置成y的祖先
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parent[i] = rootY;
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}
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}
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}
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}
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