代码随想录Day70（图论Part06）

108.冗余连接

题目：108. 冗余连接 (kamacoder.com)

思路：每次更新输出的边，来保证删除的是输入中最后出现的那条边。关键是，我要知道哪条边可以删除，而且是在join的时候就判断

尝试（难得AC）

import java.util.Scanner;public class Main {private static int[] father;private static int s1 =0;private static int t1 =0;public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt(); // 节点数量// 初始化并查集father = new int[n + 1];init(n);// 读取边并构建并查集for (int i = 0; i < n; i++) {int s = scanner.nextInt();int t = scanner.nextInt();join(s, t);}System.out.println(s1+" "+t1);}// 并查集初始化private static void init(int n) {for (int i = 1; i <= n; i++) {father[i] = i;}}// 并查集里寻根的过程private static int find(int u) {if (u != father[u]) {father[u] = find(father[u]);}return father[u];}// 判断 u 和 v 是否找到同一个根private static boolean isSame(int u, int v) {return find(u) == find(v);}// 将 v -> u 这条边加入并查集private static void join(int u, int v) {int rootU = find(u);int rootV = find(v);if (rootU != rootV) {father[rootV] = rootU;}else{s1 = u;t1 = v;}}
}

小结

基于【寻找存在的路径】代码改造，如果发现输入的（s，t）指向同一个根，说明是冗余连接，通过s1，t1每次join的时候更新

109.冗余连接||

题目：109. 冗余连接II (kamacoder.com)

思路：按照题目的意思，至少有一条边是可以删除的，我想通过fa数组，找到fa中最少被指向的元素，删掉该边，或者是说，输出该边

在遍历father数组时，还需要记录是哪条边，或许可以用set来存储

尝试（标题4）

import java.util.*;class Main {public static int[] father;public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();father = new int[n + 1];init(n);for (int i = 0; i < n; i++) { // 修正循环范围int s = scanner.nextInt();int t = scanner.nextInt();join(s, t);}int[] count = new int[n + 1];Map<Integer, Integer> map = new HashMap<>(); // 使用 Mapfor (int i = 1; i <= n; i++) { // 修正循环范围int root = find(i); // 确保父节点是根节点count[root]++;map.put(i, root);}for (int i = 1; i <= n; i++) { // 修正循环范围if (count[i] == 1) {System.out.println(map.get(i) + " " + i);}}}public static void init(int n) {for (int i = 1; i <= n; i++) {father[i] = i;}}public static int find(int u) {if (u != father[u]) {father[u] = find(father[u]);}return father[u];}public static void join(int u, int v) {int rootU = find(u);int rootV = find(v);if (rootU != rootV) {father[rootU] = rootV;}}
}

答案

import java.util.*;public class Main {public static int n;public static int[] father;public static int[] inDegree;public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();father = new int[n + 1];inDegree = new int[n + 1];List<int[]> edges = new ArrayList<>();for (int i = 0; i < n; i++) {int s = scanner.nextInt();int t = scanner.nextInt();inDegree[t]++;edges.add(new int[]{s, t});}List<Integer> vec = new ArrayList<>(); // 记录入度为2的边（如果有的话就两条边）// 找入度为2的节点所对应的边，注意要倒序，因为优先删除最后出现的一条边for (int i = n - 1; i >= 0; i--) {if (inDegree[edges.get(i)[1]] == 2) {vec.add(i);}}if (!vec.isEmpty()) {// 放在vec里的边已经按照倒序放的，所以这里就优先删vec.get(0)这条边if (isTreeAfterRemoveEdge(edges, vec.get(0))) {System.out.println(edges.get(vec.get(0))[0] + " " + edges.get(vec.get(0))[1]);} else {System.out.println(edges.get(vec.get(1))[0] + " " + edges.get(vec.get(1))[1]);}return;}// 处理情况三// 明确没有入度为2的情况，那么一定有有向环，找到构成环的边返回就可以了getRemoveEdge(edges);}// 并查集初始化public static void init() {for (int i = 1; i <= n; ++i) {father[i] = i;}}// 并查集里寻根的过程public static int find(int u) {return u == father[u] ? u : (father[u] = find(father[u]));}// 将v->u 这条边加入并查集public static void join(int u, int v) {u = find(u);v = find(v);if (u == v) return;father[v] = u;}// 判断 u 和 v是否找到同一个根public static boolean same(int u, int v) {u = find(u);v = find(v);return u == v;}// 在有向图里找到删除的那条边，使其变成树public static void getRemoveEdge(List<int[]> edges) {init(); // 初始化并查集for (int i = 0; i < n; i++) { // 遍历所有的边if (same(edges.get(i)[0], edges.get(i)[1])) { // 构成有向环了，就是要删除的边System.out.println(edges.get(i)[0] + " " + edges.get(i)[1]);return;} else {join(edges.get(i)[0], edges.get(i)[1]);}}}// 删一条边之后判断是不是树public static boolean isTreeAfterRemoveEdge(List<int[]> edges, int deleteEdge) {init(); // 初始化并查集for (int i = 0; i < n; i++) {if (i == deleteEdge) continue;if (same(edges.get(i)[0], edges.get(i)[1])) { // 构成有向环了，一定不是树return false;}join(edges.get(i)[0], edges.get(i)[1]);}return true;}
}