AGC 020 F Arcs on a Circle 题解

AGC 020 F Arcs on a Circle 题解

算法: 暴力枚举，dp

考虑端点只在整点的情况，可以破环为链，然后bitmask dp。

不在整点的时候只需要枚举大小关系的全排列即可。

code:

/*
{
######################
#       Author       #
#        Gary        #
#        2021        #
######################
*/
#include<bits/stdc++.h>
#define rb(a,b,c) for(int a=b;a<=c;++a)
#define rl(a,b,c) for(int a=b;a>=c;--a)
#define LL long long
#define IT iterator
#define PB push_back
#define II(a,b) make_pair(a,b)
#define FIR first
#define SEC second
#define FREO freopen("check.out","w",stdout)
#define rep(a,b) for(int a=0;a<b;++a)
#define SRAND mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define random(a) rng()%a
#define ALL(a) a.begin(),a.end()
#define POB pop_back
#define ff fflush(stdout)
#define fastio ios::sync_with_stdio(false)
#define check_min(a,b) a=min(a,b)
#define check_max(a,b) a=max(a,b)
using namespace std;
//inline int read(){
//    int x=0;
//    char ch=getchar();
//    while(ch<'0'||ch>'9'){
//        ch=getchar();
//    }
//    while(ch>='0'&&ch<='9'){
//        x=(x<<1)+(x<<3)+(ch^48);
//        ch=getchar();
//    }
//    return x;
//}
const int INF=0x3f3f3f3f;
typedef pair<int,int> mp;
/*}
*/
const int MAXN=6;
double rest=0;
int n,c,l[MAXN],p[MAXN],id[MAXN];
double dp[51][1<<MAXN][MAXN][51];
bool cmp(int A,int B){
	return p[A]<p[B];
}
int main(){
	scanf("%d%d",&n,&c);
	rep(i,n) scanf("%d",&l[i]);
	sort(l,l+n);
	int tmpl=l[n-1];
	--n;
	rep(i,n) p[i]=i;
	p[n]=-1;
	do{
		memset(dp,0,sizeof(dp));
		dp[0][0][n][tmpl]=1;
		rep(i,n) id[i]=i;
		sort(id,id+n,cmp);
		rb(i,0,c-1){
			rep(pt_,n){
				int pt=id[pt_];
				rep(mask,1<<n){
					rep(j,n+1){
						rep(k,c+1){
							if(i>k) continue;
							if(dp[i][mask][j][k]>0){
								if((mask>>pt)&1);
								else{
									if(i==k&&(p[j]<p[pt])) continue;
									int to=min(c,i+l[pt]);
									if(II(to,p[pt])>II(k,p[j])){
										dp[i][mask|(1<<pt)][pt][to]+=dp[i][mask][j][k];
									}
									else{
										dp[i][mask|(1<<pt)][j][k]+=dp[i][mask][j][k];
									}
								}
							}
						}
					}
				}
			}
			rep(mask,1<<n){
				rep(j,n+1){
					rep(k,c+1){
						if(i>k) continue;
						if(dp[i][mask][j][k]>0){
							dp[i+1][mask][j][k]+=dp[i][mask][j][k];
						}
					}
				}
			}
		}
		rb(j,0,n){
			rest+=dp[c][(1<<n)-1][j][c];
		}
	}while(next_permutation(p,p+n));
	rb(i,1,n) rest/=i,rest/=c;
	printf("%.16f\n",rest);
	return 0;
}