poj 1037 三维dp
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 7221 | Accepted: 2723 |
Description
A wooden fence consists of N wooden planks, placed vertically in a row next to each other. A fence looks cute if and only if the following conditions are met:
�The planks have different lengths, namely 1, 2, . . . , N plank length units.
�Each plank with two neighbors is either larger than each of its neighbors or smaller than each of them. (Note that this makes the top of the fence alternately rise and fall.)
It follows, that we may uniquely describe each cute fence with N planks as a permutation a1, . . . , aN of the numbers 1, . . . ,N such that (any i; 1 < i < N) (ai − ai−1)*(ai − ai+1) > 0 and vice versa, each such permutation describes a cute fence.
It is obvious, that there are many dierent cute wooden fences made of N planks. To bring some order into their catalogue, the sales manager of ACME decided to order them in the following way: Fence A (represented by the permutation a1, . . . , aN) is in the catalogue before fence B (represented by b1, . . . , bN) if and only if there exists such i, that (any j < i) aj = bj and (ai < bi). (Also to decide, which of the two fences is earlier in the catalogue, take their corresponding permutations, find the first place on which they differ and compare the values on this place.) All the cute fences with N planks are numbered (starting from 1) in the order they appear in the catalogue. This number is called their catalogue number.
After carefully examining all the cute little wooden fences, Richard decided to order some of them. For each of them he noted the number of its planks and its catalogue number. Later, as he met his friends, he wanted to show them the fences he ordered, but he lost the catalogue somewhere. The only thing he has got are his notes. Please help him find out, how will his fences look like.
Input
Each of the following K lines contains two integers N and C (1 <= N <= 20), separated by a space. N is the number of planks in the fence, C is the catalogue number of the fence.
You may assume, that the total number of cute little wooden fences with 20 planks fits into a 64-bit signed integer variable (long long in C/C++, int64 in FreePascal). You may also assume that the input is correct, in particular that C is at least 1 and it doesn抰 exceed the number of cute fences with N planks.
Output
Sample Input
2
2 1
3 3
Sample Output
1 2
2 3 1
Source
/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#define ll long long
#define mod 1000000007
#define PI acos(-1.0)
using namespace std;
int UP=;
int DOWN=;
ll c[][][];
void init(int n)
{
memset(c,,sizeof(c));
c[][][UP]=c[][][DOWN]=;//初始为1种
for(int i=; i<=n; i++)
{
for(int k=; k<=i; k++)
{
for(int m=k; m<i; m++)
c[i][k][UP]+=c[i-][m][DOWN];//前i-1的down方案m>=k
for(int l=; l<=k-; l++)
c[i][k][DOWN]+=c[i-][l][UP];//前i-1的up方案 l<k
}
}
}
void fun(int n,ll cc)//排序计数处理 ,一位一位的判断 不断靠近cc
{
ll skipped=;//已经跳过的方案数
int seq[];
int used[];
memset(used,,sizeof(used));
for(int i=; i<=n; i++)
{
ll oldval=skipped;
int k;
int no=;
for(k=; k<=n; k++)
{
oldval=skipped;
if(!used[k])
{
no++;//k是剩下的木棒里第no短的
if(i==)//首位
skipped+=c[n][no][UP]+c[n][no][DOWN];
else
{ //剩下n-i+1条木棒 现在放置第no短的木棒k 判断k与已经确定的seq的前一条木棒的关系
if(k>seq[i-]&&(i<=||seq[i-]>seq[i-]))
skipped+=c[n-i+][no][DOWN];//判断合理的放置
else if(k<seq[i-]&&(i<=||seq[i-]<seq[i-]))
skipped+=c[n-i+][no][UP];
}
if(skipped>=cc)//当跳过的方案数大于询问的数目跳出
break;
}
}
used[k]=;
seq[i]=k;
skipped=oldval;
}
for(int i=; i<=n; i++)
if(i<n) printf("%d ",seq[i]);
else printf("%d\n",seq[i]); }
int main()
{
int T,s;
ll c;
init();
scanf("%d",&T);
while(T--)
{
scanf("%d %I64d",&s,&c);
fun(s,c);
}
return ;
}
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