本文主要是介绍图的割点、割线问题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
图的割点问题
邻接矩阵表示图
package com.hyl.algorithm.other;import java.util.Arrays;import com.hyl.algorithm.search.base.SearchIntf;
import com.hyl.algorithm.search.base.SearchIntfFactory;/*** 寻找图的割点* <p>** @author Hyl* @version V 0.1* @since 0.1 2020-07-01 05:15*/
public class CutPoint implements SearchIntf {private ArrayGraph graph;private int n, m, root, start, timestamp;private int[] num;private int[] low;private int[] book;@Overridepublic void init() {// 图// 1// / \// 4 3// \ /// 2// / \// 5 - 6n = 6;m = 7;graph = new ArrayGraph(n);graph.addUndirected(1, 3, 1);graph.addUndirected(1, 4, 1);graph.addUndirected(2, 3, 1);graph.addUndirected(2, 4, 1);graph.addUndirected(2, 5, 1);graph.addUndirected(2, 6, 1);graph.addUndirected(5, 6, 1);num = new int[n + 1];low = new int[n + 1];book = new int[n + 1];root = start = 1;timestamp = 0;}@Overridepublic void find() {dfs(start, root);}/*** 深度优先--割点算法* @param cur 当前结点* @param father 父结点*/private void dfs(int cur, int father) {// 儿子数int child = 0;// 时间戳累加timestamp++;// 登记时间戳num[cur] = timestamp;low[cur] = timestamp;// 遍历连通结点for (int i = 1; i <= 6; i++) {// 判断连通if (graph.getPath(cur, i) == 1) {// 时间戳等于0表示没有被访问过if (num[i] == 0) {// 确定父子关系，儿子数加一child++;// 递归dfs(i, cur);// 递归完成，树已经形成，子结点和当前结点中使用能访问的较小时间戳low[cur] = Math.min(low[cur], low[i]);// 判定是否是割点，如果不是根结点，子结点能访问最小时间戳大于等于当前时间戳if (cur != root && low[i] >= num[cur]) {book[cur] = 1;}// 如果是根结点有两个儿子，就是可以确定是割点if (cur == root && child == 2) {book[cur] = 1;}} else if (i != father) {// 没有搜索到就已经有时间戳了，还不是父结点，说明在没有经过父结点就已经访问过，使用最小时间戳low[cur] = Math.min(low[cur], num[i]);}}}}@Overridepublic void print() {graph.print();System.out.println(Arrays.toString(num));System.out.println(Arrays.toString(low));System.out.println(Arrays.toString(book));}public static void main(String[] args) {SearchIntfFactory.produce(CutPoint.class);}}
图的割线问题
邻接表表示图
package com.hyl.algorithm.other;import java.util.Arrays;import com.hyl.algorithm.search.base.SearchIntf;
import com.hyl.algorithm.search.base.SearchIntfFactory;/*** 图的割边问题* <p>** @author Hyl* @version V 0.1* @since 0.1 2020-07-01 12:07*/
public class CutLine implements SearchIntf {private LinkedGraph graph;private int n, m, start, root, timestamp;private int[] num, low;private int[][] lines;private int size;@Overridepublic void init() {// 图// 1// / \// 4 3// \ /// 2// / // 5 - 6n = 6;m = 6;graph = new LinkedGraph(n, m);graph.addUndirected(1, 3, 1);graph.addUndirected(1, 4, 1);graph.addUndirected(2, 3, 1);graph.addUndirected(2, 4, 1);graph.addUndirected(2, 5, 1);graph.addUndirected(5, 6, 1);num = new int[n + 1];low = new int[n + 1];lines = new int[m][2];size = 0;root = start = 1;timestamp = 0;}@Overridepublic void find() {dfs(start, root);}private void dfs(int cur, int father) {timestamp++;num[cur] = timestamp;low[cur] = timestamp;int index = graph.first[cur];while (index != -1) {// 没有被访问过if (num[graph.v[index]] == 0) {dfs(graph.v[index], cur);low[cur] = Math.min(low[cur], low[graph.v[index]]);// 相对割点问题，>= 改为 >，表示连父节点也无法返回if (low[graph.v[index]] > num[cur]) {lines[size][0] = cur;lines[size][1] = graph.v[index];size++;}} else if (graph.v[index] != father) {// 可以被访问到，并不是父结点low[cur] = Math.min(low[cur], num[graph.v[index]]);}index = graph.next[index];}}@Overridepublic void print() {graph.printLines();System.out.println(Arrays.toString(num));System.out.println(Arrays.toString(low));for (int i = 0; i < size; i++) {System.out.print("(" + lines[i][0] + "," + lines[i][1] + ")\t");}System.out.println();}public static void main(String[] args) {SearchIntfFactory.produce(CutLine.class);}}这篇关于图的割点、割线问题的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
