POJ 1095 Trees Made to Order（计数问题）

标签：dp

Description

We can number binary trees using the following scheme: 
The empty tree is numbered 0. 
The single-node tree is numbered 1. 
All binary trees having m nodes have numbers less than all those having m+1 nodes. 
Any binary tree having m nodes with left and right subtrees L and R is numbered n such that all trees having m nodes numbered > n have either Left subtrees numbered higher than L, or A left subtree = L and a right subtree numbered higher than R. 

The first 10 binary trees and tree number 20 in this sequence are shown below: 

Your job for this problem is to output a binary tree when given its order number. 

Input

Output

Sample Input

1
20
31117532
0

Sample Output

X
((X)X(X))X
(X(X(((X(X))X(X))X(X))))X(((X((X)X((X)X)))X)X)

Source

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i < b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 20

int dp[N];

void inint()
{
	int i,j;
	dp[0]=dp[1]=1;
	for(i=2;i<N;i++)
	{
		fre(j,0,i)
		  dp[i]+=dp[j]*dp[i-1-j];  //i个节点的情况有列举左边放j，因为根节点需要一个，那么右边i-1-j
	}

}

void dfs(int n,int k)    //n个节点,第k大
{
	if(n==1)    //开始是k==1,一直错
	{
		pf("X");
		return ;
	}

	int i=0;
    for(i=0;;i++)
		if(dp[i]*dp[n-i-1]>=k) break;  //左边有dp[i]种情况,右边有dp[n-i-1],因为有一个点当根节点
	 else  k-=dp[i]*dp[n-i-1];

    if(i)       //左边有节点
	{
		pf("(");
		dfs(i,(k-1)/dp[n-i-1]+1);
		pf(")");
	}
	pf("X");
	if(n-i-1)    //右边有节点
	{
		pf("(");
		dfs(n-1-i,(k-1)%dp[n-1-i]+1);
		pf(")");
	}
}
int main()
{
	int i,j,n;
	inint();
	while(sf(n),n)
	{
		for(i=1;;i++)
		  if(dp[i]>=n) break; //已经锁定i个节点是满足条件的
		 else n-=dp[i];
		dfs(i,n);
      pf("\n");
	}
    return 0;
}

POJ 1095 Trees Made to Order（计数问题）

标签：dp

原文地址：http://blog.csdn.net/u014737310/article/details/45566277