tyvj 2002 扑克牌

期望DP

本题递推比较麻烦,可以记忆化搜索

注意搜索的边界条件

以及每一次转移

#include <iostream>

#include <cstdio>

#include <cmath>

#include <algorithm>

#include <cstring>

using namespace std;

double dp[16][16][16][16][6][6];

bool f[16][16][16][16][6][6];

int term[4];

double dfs(int a, int b, int c, int d, int p, int q) {

	if(f[a][b][c][d][p][q]) return dp[a][b][c][d][p][q];

	f[a][b][c][d][p][q] = 1;

	double & ans = dp[a][b][c][d][p][q];ans = 0.0;

	int r1 = a, r2 = b, r3 = c, r4 = d;

	if(p == 1) r1++; if(p == 2) r2++; if(p == 3) r3++; if(p == 4) r4++;

	if(q == 1) r1++; if(q == 2) r2++; if(q == 3) r3++; if(q == 4) r4++;

	if(r1 >= term[0] && r2 >= term[1] && r3 >= term[2] && r4 >= term[3]) return ans = 0;

	int sum = 54 - r1 - r2 - r3 - r4;

	if(sum <= 0) return ans = 1e10;

	if(a < 13) ans += dfs(a + 1, b, c, d, p, q) * (13 - a) /sum;

	if(b < 13) ans += dfs(a, b + 1, c, d, p, q) * (13 - b) /sum;

	if(c < 13) ans += dfs(a, b, c + 1, d, p, q) * (13 - c) /sum;

	if(d < 13) ans += dfs(a, b, c, d + 1, p, q) * (13 - d) /sum;

	if(!p) {

		double tmp = dfs(a, b, c, d, 1, q);

		tmp = min(tmp, dfs(a, b, c, d, 2, q));

		tmp = min(tmp, dfs(a, b, c, d, 3, q));

		tmp = min(tmp, dfs(a, b, c, d, 4, q));

		ans += tmp / sum;

	}

	if(!q) {

		double tmp = dfs(a, b, c, d, p, 1);

		tmp = min(tmp, dfs(a, b, c, d, p, 2));

		tmp = min(tmp, dfs(a, b, c, d, p, 3));

		tmp = min(tmp, dfs(a, b, c, d, p, 4));

		ans += tmp / sum;

	}

	ans += 1.0;//概率和是 1 ,所以我们直接加 1 即可

	return ans;

}

int main() {

	for(int i = 0; i < 4; i++) cin >> term[i];

	double ans = dfs(0, 0, 0, 0, 0, 0);

	if(ans >= 100.0) printf("-1.000\n");

	else printf("%.3f\n", ans);

	return 0;

}

巴特西

tyvj 2002 扑克牌

期望DP

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