本文主要是介绍ACMjava无根树转有根数，构建表达式，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！
package com.supermars.practice;import java.util.Arrays;
import java.util.Scanner;
import java.util.Vector;public class 无根树转有根树 {static Scanner cin = new Scanner(System.in);static Vector<Integer> G[] = new Vector[1 << 7];static int p[];;public static void main(String[] args) {while (cin.hasNext()) {int n = cin.nextInt();p = new int[n];for (int i = 0; i < n - 1; i++) { // n-1条边int u = cin.nextInt();int v = cin.nextInt();if (G[u] == null)G[u] = new Vector<Integer>();G[u].add(v);if (G[v] == null)G[v] = new Vector<Integer>();G[v].add(u);}Arrays.fill(p, -1);dfs(1, -1);// 节点1开始为跟,-1表示无父节点// System.out.println(Arrays.toString(p));}}private static void dfs(int u, int fa) {int d = G[u].size();// 多少个u节点的相邻节点for (int i = 0; i < d; i++) {int v = G[u].get(i); // u的相邻节点iif (v != fa) {dfs(v, p[v] = u); // 保存v节点的父亲uSystem.out.println(v + "的父节点" + p[v]);}}}
}
package com.supermars.practice;import java.util.Scanner;public class 表达式树 {static Scanner cin = new Scanner(System.in);static final int MAXN = 1 << 7;static char input[]; // 2+3*(4-1)-5/1 表达式字符串static char op[] = new char[MAXN]; // 节点中的操作符static int lch[] = new int[MAXN]; // 每个节点的左右儿子static int rch[] = new int[MAXN];static int nc = 0;// 节点个数public static void main(String[] args) {while (cin.hasNext()) {input = cin.nextLine().toCharArray();bulidTree(input, 0, input.length);System.out.println(transTree(0));}}private static int bulidTree(char[] s, int x, int y) {int c1 = -1, c2 = -1, p = 0;int u;if (y - x == 1) { // 只剩余一个操作字符节点u = nc++;lch[u] = rch[u] = 0; // 操作字符无左右子树op[u] = s[x]; // 保存操作字符return u;}// c1 最后+- c2最后*/的位置for (int i = x; i < y; i++) {switch (s[i]) {case '(':p++;break;case ')':p--;break;case '+':case '-':if (p == 0) // 在括号外c1 = i; // 最后+-的位置break;case '*':case '/':if (p == 0) // 最后*/的位置c2 = i;break;}}if (c1 < 0)c1 = c2; // 括号外没有加减if (c1 < 0)return bulidTree(s, x + 1, y - 1);// 求解[x+1,y-1]构造表达式u = nc++; // c1划分左右子树,lch[u] = bulidTree(s, x, c1); // 构造左子树rch[u] = bulidTree(s, c1 + 1, y);op[u] = s[c1]; // 存节点操作字符return u;}private static int transTree(int cur) {if (lch[cur] == 0 || rch[cur] == 0) { // 叶子节点返回值return op[cur] - '0';} else {int retl = transTree(lch[cur]); // 计算左子树的值int retr = transTree(rch[cur]);int ret = 0;switch (op[cur]) { // 根据当前节点分别计算左右子树+-*/的结果case '+':ret = (retl + retr);break;case '-':ret = (retl - retr);break;case '*':ret = (retl * retr);break;case '/':ret = (retl / retr);break;}// System.out.println(ret);return ret;}}
}
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