Siwei Jia sub binary tree dp + (Los Valley P1040)

Plus binary tree

Title Description

A set of n nodes of the binary tree in preorder to (1,2,3, ..., n), where the number 1,2,3, ..., n is a node number. Each node has a score (both positive integers), referred to the i th fraction DI nodes, each sub-tree, and it has a plus tree, any subtree subtree (also contains tree itself) plus min is calculated as follows:

Of the left subtree of the partition subtree × subtree of the right subtree plus + fraction of root subtree.

If a sub-tree is empty, the provisions of its plus 1 divided leaves scores extra points is a leaf node itself. Irrespective of its empty tree.

Determine a preorder conform to (1,2,3, ..., n) and the highest points of the binary tree. Required output;

(1) tree of the highest points

(2) tree in preorder traversal

Input Format

Line 1: an integer n (n <30), is the number of nodes.

Line 2: n integers separated by spaces, each node as a fraction (fraction <100).

Output Format

Line 1: an integer, the highest points (Ans≤4,000,000,000).

Line 2: n integers separated by spaces, for preorder traversal of the tree.

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This question is particularly feeling thinking;

First know know preorder traversal is impossible to obtain a binary tree before and after, since only then can know preorder any point when the root;

That is the most use of this particular nature, can write equation: dp [l] [r] = max {dp [l] [k-1], dp [k + 1] [r]}

dp [l] [r] denotes the R & lt preorder l (l will be less than the given topic r) to root node k, l is a left subtree root, r is the maximum right subtree root node Add points;

This time there is a doubt if l> r how to do, there is no obvious link between the l and r on behalf of this, you can return 1;

If l == r it? L This represents a leaf node, return W [l] to;

There are prior to a traversal, ideas, as long as the recording root [l] [r] can be a root;

Code:

#include<bits/stdc++.h>
#define LL long long
#define pa pair<int,int>
#define ls k<<1
#define rs k<<1|1
#define inf 0x3f3f3f3f
using namespace std;
const int N=100010;
const int M=2000100;
const LL mod=1000007;
int n,w[100],root[100][100];
LL dp[100][100];
LL dfs1(int l,int r){
	if(l>r) return 1ll; 
	if(dp[l][r]!=-1) return dp[l][r];
	LL mmax=0;
	for(int k=l;k<=r;k++){
		LL sum=dfs1(l,k-1)*dfs1(k+1,r)+w[k];
		if(sum>mmax){
			root[l][r]=k;
			mmax=sum;
		}
	}
	return dp[l][r]=mmax;
}
void dfs2(int l,int r){
	if(l>r) return;
	int rt=root[l][r];
	cout<<rt<<" ";
	dfs2(l,rt-1);
	dfs2(rt+1,r);
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++) cin>>w[i];
	memset(dp,-1,sizeof(dp));
	for(int i=1;i<=n;i++) dp[i][i]=w[i],root[i][i]=i;
	cout<<dfs1(1,n)<<endl;
	dfs2(1,n);
	return 0;
}