P1040 + Binary Tree, p1040 + Binary Tree

P1040 + Binary Tree, p1040 + Binary Tree
Description

Set the central traversal of a tree with n nodes to (1, 2, 3 ,..., N), where the number is 1, 2, 3 ,..., N is the node number. Each node has a score (all positive integers). Note that the score of node I is di. The tree and each of its Subtrees have a plus score, the method for calculating the extra points of any subtree (also including the tree itself) is as follows:

Plus points for the left subtree of the subtree × plus points for the right subtree of the subtree + scores for the root of the subtree.

If a subtree is empty, set it to 1. The leaf score is the score of the leaf node. Ignore its empty subtree.

Try to find a tree that matches the ordinal traversal (, 3 ,..., N) the tree with the highest bonus points. Output required;

(1) Top bonus points for tree

(2) tree pre-order traversal

Input/Output Format Input Format:

Row 1st: an integer n (n <30), indicating the number of nodes.

Row 2nd: n integers separated by spaces, which are the scores of each node (score <100 ).

Output Format:

Row 1st: an integer that is the maximum value (the result cannot exceed 4,000,000,000 ).

Row 2nd: n integers separated by spaces, traversing the tree in the forward order.

Input and Output sample Input example #1:

55 7 1 2 10

Output sample #1:

1453 1 2 4 5

Interval DP.
Recurrence won't write, and then write memory will search...

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 const int MAXN=51; 7 int n,zx[MAXN]; 8 int dp[MAXN][MAXN]; 9 int root[MAXN][MAXN];10 void read(int & n)11 {12     char c='+';int x=0;bool flag=0;13     while(c<'0'||c>'9')14     {c=getchar();if(c=='-')flag=1;}15     while(c>='0'&&c<='9')16     {x=x*10+(c-48);c=getchar();}17     flag==1?n=-x:n=x;18 }19 int M_s(int l,int r)20 {21     dp[l][r]=1;22     if(l==r)23     {24         dp[l][r]=zx[l];25         root[l][r]=l;26         return zx[l];27     }28     else for(int k=l;k<=r;k++)29     {30         int lson=1,rson=1;31         if(dp[l][k-1])32             lson=dp[l][k-1];33         else if(l<=k-1)34             lson=M_s(l,k-1);35         if(dp[k+1][r])36             rson=dp[k+1][r];37         else if(r>k)38             rson=M_s(k+1,r);39         if(lson*rson+zx[k]>dp[l][r])40         {41             dp[l][r]=lson*rson+zx[k];42             root[l][r]=k;43         }44     }45     return dp[l][r];46 }47 void xianxu(int l,int r)48 {49     if(root[l][r])50     {51         printf("%d ",root[l][r]);52         xianxu(l,root[l][r]-1);53         xianxu(root[l][r]+1,r);54     }55 }56 int main()57 {58     read(n);59     for(int i=1;i<=n;i++)60         read(zx[i]);61     int out=M_s(1,n);62     printf("%d\n",out);63     xianxu(1,n);64     return 0;65 }