图的割点、割线问题

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; @Override public void init() { // 图 // 1 // / \ // 4 3 // \ / // 2 // / \ // 5 - 6 n = 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; } @Override public 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]); } } } } @Override public 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; @Override public void init() { // 图 // 1 // / \ // 4 3 // \ / // 2 // / // 5 - 6 n = 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; } @Override public 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]; } } @Override public 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); } }