Codeforces Round #551 (Div. 2)D. Serval and Rooted Tree

D. Serval and Rooted Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Now Serval is a junior high school student in Japari Middle School, and he is still thrilled on math as before.

As a talented boy in mathematics, he likes to play with numbers. This time, he wants to play with numbers on a rooted tree.

A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a node vv is the last different from vv vertex on the path from the root to the vertex vv. Children of vertex vv are all nodes for which vv is the parent. A vertex is a leaf if it has no children.

The rooted tree Serval owns has nn nodes, node 11 is the root. Serval will write some numbers into all nodes of the tree. However, there are some restrictions. Each of the nodes except leaves has an operation maxmax or minmin written in it, indicating that the number in this node should be equal to the maximum or minimum of all the numbers in its sons, respectively.

Assume that there are kk leaves in the tree. Serval wants to put integers 1,2,…,k1,2,…,k to the kk leaves (each number should be used exactly once). He loves large numbers, so he wants to maximize the number in the root. As his best friend, can you help him?

Input

The first line contains an integer n (2≤n≤3⋅1052≤n≤3⋅105), the size of the tree.

The second line contains nn integers, the ii-th of them represents the operation in the node ii. 00 represents minmin and 11 represents maxmax. If the node is a leaf, there is still a number of 00 or 11, but you can ignore it.

The third line contains n−1n−1 integers f2,f3,…,fnf2,f3,…,fn (1≤fi≤i−11≤fi≤i−1), where fifi represents the parent of the node ii.

Output

Output one integer — the maximum possible number in the root of the tree.

Examples

input

Copy

6
1 0 1 1 0 1
1 2 2 2 2

output

Copy

1

input

Copy

5
1 0 1 0 1
1 1 1 1

output

Copy

4

input

Copy

8
1 0 0 1 0 1 1 0
1 1 2 2 3 3 3

output

Copy

4

input

Copy

9
1 1 0 0 1 0 1 0 1
1 1 2 2 3 3 4 4

output

Copy

5

Note

Pictures below explain the examples. The numbers written in the middle of the nodes are their indices, and the numbers written on the top are the numbers written in the nodes.

In the first example, no matter how you arrange the numbers, the answer is 11.

In the second example, no matter how you arrange the numbers, the answer is 44.

In the third example, one of the best solution to achieve 44 is to arrange 44 and 55 to nodes 44 and 55.

In the fourth example, the best solution is to arrange 55 to node 55.

 

官方题解：

1153D - Serval and Rooted Tree

Author & preparation: bzh

Editorial

If we want to check whether x is the answer (I didn't say I want to do binary search), then we can set all the numbers no less than x as 1, and the numbers less than x as 0. Then we can use dpidpi to represent that, there should be at least dpidpi ones in the subtree of ii such that the number on ii is one. Then k+1−dp1 is the final answer. Complexity O(n).

叶子节点dp[i]=1

如果节点取max则dp[i]=min(dp[子节点们])

如果取min则dp[i]+=dp[子节点们]

答案就是  叶子节点个数+1-dp[1]

#include<bits/stdc++.h>
#include<vector>
using namespace std;
vector<int> g[300005];
int dp[300005];
int a[300005];
map<int,int>vis;
void dfs(int now) {

	if(g[now].size()==0) {
		dp[now]=1;
		return ;
	}
	if(a[now]==1) {
		dp[now]=0x3f3f3f3f;
		for(int i=0; i<g[now].size(); i++) {
			dfs(g[now][i]);
			dp[now]=min(dp[now],dp[g[now][i]]);
		}
	} else {
		for(int i=0; i<g[now].size(); i++) {
			dfs(g[now][i]);
			dp[now]+=dp[g[now][i]];
		}
	}
	return ;
}
int main() {
	int n;
	cin>>n;
	int cnt=n;
	for(int i=1; i<=n; i++) {
		cin>>a[i];
	}

	for(int i=2; i<=n; i++) {
		int fa;
		cin>>fa;
		if(!vis[fa]) {
			cnt--,vis[fa]=1;
		}
		g[fa].push_back(i);
	}
	dfs(1);
	cout<<cnt+1-dp[1];
	return 0;
}