[HDOJ5058]So easy

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5058

 

强行二叉排序树(0ms)(stl set可做(31ms))

 

 1 #pragma warning(disable:4996)
 2 
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <iomanip>
 6 #include <cstring>
 7 #include <climits>
 8 #include <complex>
 9 #include <fstream>
10 #include <cassert>
11 #include <cstdio>
12 #include <bitset>
13 #include <vector>
14 #include <deque>
15 #include <queue>
16 #include <stack>
17 #include <ctime>
18 #include <set>
19 #include <map>
20 #include <cmath>
21 
22 using namespace std;
23 
24 typedef struct Node {
25     Node* left;
26     Node* right;
27     int data;
28     Node() { left = NULL; right = NULL; }
29 }Node;
30 
31 const int maxn = 110;
32 Node memory[maxn << 1];
33 int CNT;
34 int n;
35 int tmp;
36 int cnt;
37 int a[maxn], b[maxn];
38 
39 Node* insert(Node* cur, int data) {
40     if (cur == NULL) {
41         cur = &memory[CNT++];
42         cur->data = data;
43         return cur;
44     }
45     if (data > cur->data) {
46         cur->right = insert(cur->right, data);
47     }
48     if(data < cur->data){
49         cur->left = insert(cur->left, data);
50     }
51     return cur;
52 }
53 
54 void inorder(int* x, Node* root) {
55     if (root) {
56         inorder(x, root->left);
57         x[cnt++] = root->data;
58         inorder(x, root->right);
59     }
60 }
61 
62 int main() {
63     //freopen("input", "r", stdin);
64     while (~scanf("%d", &n)) {
65         CNT = 0;
66         memset(a, 0, sizeof(a));
67         memset(b, 0, sizeof(b));
68         memset(memory, 0, sizeof(memory));
69         Node* root1 = NULL;
70         for (int i = 0; i < n; i++) {
71             scanf("%d", &tmp);
72             root1 = insert(root1, tmp);
73         }
74         Node* root2 = NULL;
75         for (int i = 0; i < n; i++) {
76             scanf("%d", &tmp);
77             root2 = insert(root2, tmp);
78         }
79         cnt = 0;
80         inorder(a, root1);
81         cnt = 0;
82         inorder(b, root2);
83         int flag = 0;
84         for (int i = 0; i < n; i++) {
85             if (a[i] != b[i]) {
86                 flag = 1;
87                 break;
88             }
89         }
90         if (flag)    printf("NO\n");
91         else        printf("YES\n");
92     }
93     return 0;
94 }

 

转载于:https://www.cnblogs.com/kirai/p/4911545.html