Codeforces1153——D. Serval and Rooted Tree（思维好题+dfs+贪心）

原题链接
D. Serval and Rooted Tree
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 v is the last different from v vertex on the path from the root to the vertex v. Children of vertex v are all nodes for which v is the parent. A vertex is a leaf if it has no children.

The rooted tree Serval owns has n nodes, node 1 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 max or min 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 k leaves in the tree. Serval wants to put integers 1,2,…,k to the k 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⋅105), the size of the tree.

The second line contains n integers, the i-th of them represents the operation in the node i. 0 represents min and 1 represents max. If the node is a leaf, there is still a number of 0 or 1, but you can ignore it.

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

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

Examples
inputCopy
6
1 0 1 1 0 1
1 2 2 2 2
outputCopy
1
inputCopy
5
1 0 1 0 1
1 1 1 1
outputCopy
4
inputCopy
8
1 0 0 1 0 1 1 0
1 1 2 2 3 3 3
outputCopy
4
inputCopy
9
1 1 0 0 1 0 1 0 1
1 1 2 2 3 3 4 4
outputCopy
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 1.

In the second example, no matter how you arrange the numbers, the answer is 4.

In the third example, one of the best solution to achieve 4 is to arrange 4 and 5 to nodes 4 and 5.

In the fourth example, the best solution is to arrange 5 to node 5.

题意：

思路：

是一个有idea但是没有写出来的题。
首先，贪心的考虑，假设对于非叶子节点$x$:
1.当该节点的操作为$max$时：该点的值为叶子节点的最大值。
2.当该节点的操作为$min$时：该点要尽可能的取到最大值。
这时候要尽可能的把大的放到叶子节点数量少的点。

比如这个图，假设$1,2,3,4$节点的值都是$min$，那么考虑如何分配是最优的：
就是把大的尽可能分给$12,13$。
实现起来就很思维了。
数组$d[i]$维护的是以$i$为根节点的子树中叶子节点的数值排名$d[i]$大的数值。
比如该节点$x$下有6个叶子节点,$d[x]=2$，表示这个节点最大可以是6-2+1=5，也就是第二大的数。
下面就是分类讨论了：
1.当前节点$x$为叶子节点,$d[x]=1$
2.当前节点$x$的值为$min$,$d[x]={\sum }_{y=so{n}_x}^{}{d}_y$
3.当前节点的值为$max$，$d[x]=min(d[y])$
对于分类讨论的内容可以画图理解。
答案为叶子节点个数+1-$d[1]$

代码：

///#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read(){ll x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')f = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p){ll res = 1;while(b){if(b & 1)res = res * a % p;a = a * a % p;b >>= 1;}return res;}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const double eps = 1e-8;
const int maxn = 3e5 + 7,mod=1000000007;
int d[maxn],a[maxn];
vector<int>g[maxn];
int n,cnt=0;
void dfs(int u){
	if(!g[u].size()){
		d[u]=1;
		cnt++;
		return ;
	}
	if(a[u]) d[u]=n;
	else d[u]=0;
	for(int t:g[u]){
		dfs(t);
		if(a[u]) d[u]=min(d[u],d[t]);
		else d[u]+=d[t];
	}
}
int main(){
	n=read;
	rep(i,1,n) a[i]=read;
	rep(i,2,n){
		int x=read;
		g[x].push_back(i);
	}
	dfs(1);
	cout<<cnt+1-d[1]<<endl;
	return 0;
}

参考