To search a key in a binary search tree, we start from the root and move all the way down, choosing branches according to the comparison results of the keys. The searching path corresponds to a sequence of keys. For example, following {1, 4, 2, 3} we can find 3 from a binary search tree with 1 as its root. But {2, 4, 1, 3} is not such a path since 1 is in the right subtree of the root 2, which breaks the rule for a binary search tree. Now given a sequence of keys, you are supposed to tell whether or not it indeed correspnds to a searching path in a binary search tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N and M (<=100) which are the total number of sequences, and the size of each sequence, respectively. Then N lines follow, each gives a sequence of keys. It is assumed that the keys are numbered from 1 to M.
Output Specification:
For each sequence, print in a line "YES" if the sequence does correspnd to a searching path in a binary search tree, or "NO" if not.
Sample Input:
3 4 1 4 2 3 2 4 1 3 3 2 4 1
Sample Output:
YES NO NO 解题思路:如果第i+1个数比i个数大（小），i+1后面的数都比i大（小）解题感悟：可以用标志位连续跳出两个for循环
1 #include <iostream>
2 #include <vector>
3 using namespace std;
4 int main(){
5 int n, m;
6 cin >> n >> m;
7 for (int i = 0; i < n;i++)
8 {
9 int flag = 0;
10 vector<int> v;
11 for (int j = 0; j < m;j++)
12 {
13 int elem;
14 cin >> elem;
15 v.push_back(elem);
16 }
17 for (int j = 0; j < m - 2;j++)
18 {
19 if (v[j]>v[j + 1])
20 {
21 for (int k = j + 2; k < m;k++)
22 {
23 if (v[k]>=v[j])
24 {
25 flag = 1;
26 break;
27 }
28 }
29 }
30 else
31 {
32 for (int k = j + 2; k < m; k++)
33 {
34 if (v[k] <= v[j])
35 {
36 flag = 1;
37 break;
38 }
39 }
40 }
41 if (flag == 1)
42 break;
43 }
44 if (flag == 1)
45 cout << "NO"<<endl;
46 else
47 cout << "YES"<<endl;
48 }
49 return 0;
50 }