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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1e6 + 10;
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//树的结构体+存储数组
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//此处:为二叉树的标准创建、遍历模板,可用于其它试题!
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struct Node {
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int id; // 当前结点ID
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int left; // 左结点ID
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int right; // 右结点ID
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char value; //当前结点的value值(本题中未使用)
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} t[N];
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//前序遍历
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void pre_order(Node node) {
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//利用递归前序输出二叉树
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if (node.id) {
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cout << node.value;
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pre_order(t[node.left]);
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pre_order(t[node.right]);
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}
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}
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int n;
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/**
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测试用例:
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6
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abc
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bdi
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cj*
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d**
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i**
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j**
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参考答案:abdicj
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*/
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int b[N];
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int main() {
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cin >> n; //二叉树的结点数量
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//读入每个结点及它的左右儿子
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for (int i = 1; i <= n; i++) {
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string str;
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cin >> str;
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int j = str[0] - 'a' + 1;//结点编号
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b[j]++;//没有父亲的才是根
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//结点编号
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t[j].id = j;
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//记录字母值
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t[j].value=str[0];
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//记录左儿子
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if (str[1] != '*') t[j].left = str[1] - 'a' + 1, b[t[j].left]++;;
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//记录右儿子
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if (str[2] != '*') t[j].right = str[2] - 'a' + 1, b[t[j].right]++;;
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}
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//第一次提交,30分,似乎要求通过运算找到根,第一个输入的未必是根
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//第二次,记录谁是根,AC。
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int root;
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for (int i = 1; i <= 26; i++)
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if (b[i] == 1) {
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root = i;
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break;
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}
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//前序遍历
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pre_order(t[root]);
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return 0;
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}
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