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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 }
Search in a Binary Search Tree
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原文地址：http://www.cnblogs.com/lkyblog/p/4422685.html
