【HDOJ6628】permutation 1（dfs）

题意：求1到n的排列中使得其差分序列的字典序为第k大的原排列

n<=20，k<=1e4

思路：爆搜差分序列，dfs时候用上界和下界剪枝

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
#define N  1100000
#define M  4100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=998244353,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;

int b[100],c[100],ans[100],d[100],n,K;

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

void dfs(int k,int l,int r)
{
    if(l>r) return;
    if(K==0) return;
    if(k==n)
    {
        K--;
        if(K==0)
        {
            rep(i,1,n) ans[i]=c[i]+l;
        }
        return;
    }
    rep(i,1-c[k]-n,n-c[k]-1)
    {
        int t=c[k]+i+n;
        if(!b[t])
        {
            b[t]=1;
            c[k+1]=c[k]+i;
            dfs(k+1,max(l,-c[k+1]+1),min(r,n-c[k+1]));
            b[t]=0;
        }
    }
}

int main()
{
    int cas=read();
    while(cas--)
    {
        n=read(),K=read();
        rep(i,1,n+n) b[i]=0;
        b[n]=1;
        dfs(1,1,n);
        rep(i,1,n-1) printf("%d ",ans[i]);
        printf("%d
",ans[n]);
    }

    return 0;
}