Educational Codeforces Round 19-D. Broken BST

原題鏈接

D. Broken BST

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let T be arbitrary binary tree — tree, every vertex of which has no more than two children. Given tree is rooted, so there exists only one vertex which doesn't have a parent — it's the root of a tree. Every vertex has an integer number written on it. Following algorithm is run on every value from the tree T:

1. Set pointer to the root of a tree.
2. Return success if the value in the current vertex is equal to the number you are looking for
3. Go to the left child of the vertex if the value in the current vertex is greater than the number you are looking for
4. Go to the right child of the vertex if the value in the current vertex is less than the number you are looking for
5. Return fail if you try to go to the vertex that doesn't exist

Here is the pseudo-code of the described algorithm:

bool find(TreeNode t, int x) {
    if (t == null)
        return false;
    if (t.value == x)
        return true;
    if (x < t.value)
        return find(t.left, x);
    else
        return find(t.right, x);
}
find(root, x);

The described algorithm works correctly if the tree is binary search tree (i.e. for each node the values of left subtree are less than the value in the node, the values of right subtree are greater than the value in the node). But it can return invalid result if tree is not a binary search tree.

Since the given tree is not necessarily a binary search tree, not all numbers can be found this way. Your task is to calculate, how many times the search will fail being running on every value from the tree.

If the tree has multiple vertices with the same values on them then you should run algorithm on every one of them separately.

Input

First line contains integer number n (1 ≤ n ≤ 105) — number of vertices in the tree.

Each of the next n lines contains 3 numbers v, l, r (0 ≤ v ≤ 109) — value on current vertex, index of the left child of the vertex and index of the right child of the vertex, respectively. If some child doesn't exist then number  - 1 is set instead. Note that different vertices of the tree may contain the same values.

Output

Print number of times when search algorithm will fail.

Examples

input

3
15 -1 -1
10 1 3
5 -1 -1

output

2

input

8
6 2 3
3 4 5
12 6 7
1 -1 8
4 -1 -1
5 -1 -1
14 -1 -1
2 -1 -1

output

1

Note

In the example the root of the tree in vertex 2. Search of numbers 5 and 15 will return fail because on the first step algorithm will choose the subtree which doesn't contain numbers you are looking for.

先根據輸入的數據構造一棵樹，然後從根節點開始對這棵樹進行深搜，初始設置maxs = -1, mins = 1e9+1, 表示maxs表示節點應該大於的最大值，mins表示節點要小於的最小值，對於某節點f的左孩子，maxs = max(maxs, f),

對於右孩子,mins = min(mins, f)，若某節點p滿足， p > maxs && p < mins 則p是能按照BST的方式搜到的

#include <bits/stdc++.h>
#define maxn 100005
#define MOD 1000000007
typedef long long ll;
using namespace std;

struct Node{
	int v, l, r, is;
}node[maxn];
int vis[maxn];
void dfs(int k, int a, int b) {
	if(node[k].v > a && node[k].v < b)
	   node[k].is = 1;
	if(node[k].l != -1)
	dfs(node[k].l, a, min(node[k].v, b));
	if(node[k].r != -1)
	dfs(node[k].r, max(a, node[k].v), b);
}
bool cmp(const Node&a, const Node&b) {
	return a.v < b.v ||(a.v == b.v && a.is < b.is);
}
int main() {
//	freopen("in.txt", "r", stdin); 
	int n, root;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) {
		scanf("%d%d%d", &node[i].v, &node[i].l, &node[i].r);
		if(node[i].l != -1)
		 vis[node[i].l] = 1;
		if(node[i].r != -1)
		 vis[node[i].r] = 1;
	}
	for(int i = 1; i <= n; i++) {
		if(vis[i] == 0) {
			root = i;
			break;
		}
	}
	int ans = 0;
	dfs(root, -1, 1000000001);
	sort(node+1, node+n+1, cmp);
	for(int i = n; i > 1; i--){
	 if(node[i].v == node[i-1].v)
	  node[i-1].is = node[i].is;
	  if(node[i].is == 0)
	   ans++;
	}
	if(node[1].is == 0)
	 ans++;
	cout << ans << endl;
	return 0;

}