LightOJ 1364 树形DP

52张扑克牌，问拿到指定数量的4个花色的最少次数期望是多少，其中拿到joker必须马上将其视作一种花色，且要使后续期望最小。

转移很容易想到，主要是两张joker的处理，一个状态除了普通的4个方向的转移，当没拿到joker时还要增加拿到joker的期望，根据题意直接在当前状态下找最小的期望计算即可。

 

/** @Date    : 2017-08-29 17:58:59
  * @FileName: LightOJ 1364 概率DP.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1 ,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

double dp[15][15][15][15][5][5];
int A, B, C, D;

double dfs(int a, int b, int c, int d, int x, int y)
{
	if(dp[a][b][c][d][x][y] > -1)
		return dp[a][b][c][d][x][y];
	double tot = a + b + c + d + (x==0?1:0) + (y==0?1:0);
	int pa = 13 - a, pb = 13 - b, pc = 13 - c, pd = 13 - d;
	if(x == y && x == 1) pa += 2;//
	else if(x == 1 || y == 1) pa++;

	if(x == y && x == 2) pb += 2;//
	else if(x == 2 || y == 2) pb++;

	if(x == y && x == 3) pc += 2;//
	else if(x == 3 || y == 3)pc++;

	if(x == y && x == 4) pd += 2;//
	else if(x == 4 || y == 4) pd++;

	if(pa >= A && pb >= B && pc >= C && pd >= D)
		return dp[a][b][c][d][x][y] = 0;
	//
	double ans = 0.0;
	if(a)
		ans += (double)a / tot * dfs(a - 1, b, c, d, x, y) * 1.000000000;
	if(b)
		ans += (double)b / tot * dfs(a, b - 1, c, d, x, y) * 1.000000000;
	if(c)
		ans += (double)c / tot * dfs(a, b, c - 1, d, x, y) * 1.000000000;
	if(d)
		ans += (double)d / tot * dfs(a, b, c, d - 1, x, y) * 1.000000000;
	if(x == 0)
	{
		double e = 0x7f;
		for(int i = 1; i <= 4; i++)
			e = min(e, dfs(a, b, c, d, i, y));
		ans += (1.00000 / tot) * e;
	}
	if(y == 0)
	{
		double e = 0x7f;
		for(int i = 1; i <= 4; i++)
			e = min(e, dfs(a, b, c, d, x, i));
		ans += (1.00000 / tot) * e;
	}
	//cout << "~" << endl;
	return dp[a][b][c][d][x][y] = ans + 1;
}
int main()
{
	int T;
	cin >> T;
	int icase = 0;
	while(T--)
	{
		scanf("%d%d%d%d", &A, &B, &C, &D);
		memset(dp, 0xc2, sizeof(dp));
		//cout << ******dp << endl;
		int cnt = (A-13>0?A-13:0) + (C-13>0?C-13:0) + (B-13>0?B-13:0) + (D-13>0?D-13:0);
		if(cnt > 2)
			printf("Case %d: -1\n", ++icase);
		else
			printf("Case %d: %.10lf\n", ++icase, dfs(13,13,13,13,0,0));

	}
    return 0;
}