AOJ1277 Minimal Backgammon - seica

問題

考察

典型的なDP
dp[i][j] :=iターン目にjにいる確率
LだったりBだったり折り返しだったりの判定に順序だけちょっと気を遣う。

#include<iostream>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<functional>
#include<stack>
#include<queue>
#include <iomanip>
#include<map>
#include<limits>
#include<cmath>
#include<algorithm>
#include<bitset>
#include<utility>
#include<complex>
#include<cstdlib>
#include<set>
#include<cctype>

#define DBG cerr << '!' << endl;
#define REP(i,n) for(int (i) = (0);(i) < (n);++i)
#define rep(i,s,g) for(int (i) = (s);(i) < (g);++i)
#define rrep(i,s,g) for(int (i) = (s);i >= (g);--(i))
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {for(int W = 0;W < (n);W++)cerr << v[W] << ' ';cerr << endl << endl;}
#define SHOW2d(v,i,j) {for(int aaa = 0;aaa < i;aaa++){for(int bbb = 0;bbb < j;bbb++)cerr << v[aaa][bbb] << ' ';cerr << endl;}cerr << endl;}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(10)

using namespace std;

typedef long long ll;
typedef vector<int> iv;
typedef vector<iv> iiv;
typedef vector<string> sv;

double mp[111][111];
bool L[111];
bool B[111];
int n,t,l,b;

int main()
{
	while(cin >> n >> t >> l >> b,n|t|l|b)
	{
		REP(i,111)REP(j,111)mp[i][j] = 0;
		REP(i,111)L[i] = B[i] = false;
		mp[0][0] = 1;
		
		REP(i,l)
		{
			int tmp;cin >> tmp;L[tmp] = true;
		}
		REP(i,b)
		{
			int tmp;cin >> tmp;B[tmp] = true;
		}
		
		REP(i,t)
		{
			REP(j,n)
			{
				if(mp[i][j] > 0)
				{
					REP(k,6)
					{
						int tmp = j + k + 1;
						int turn = i+1;
						if(tmp > n)tmp = 2*n - tmp;
						if(L[tmp])turn++;
						if(B[tmp])tmp = 0;
						mp[turn][tmp] += mp[i][j] / 6;
					}
				}
			}
		}
		
		//SHOW2d(mp,t+1,n+1);
		
		
		double ans = 0;
		REP(i,t+1)
		{
			ans += mp[i][n];
		}
		
		cout << Decimal << ans << endl;
	}

	return 0;
}