其实是水题啊,不过就是没人做= =
概率+记忆化搜索
在叠叠乐基础规则之上,已知叠叠乐中每一层只有以上四种状态是稳定的,并且在总高度为n时,积木移动成功的概率为p = b - n * d。
题意就是求在最优策略下,A的胜率。(具体一点请自行读题)
如图,只有A和C两个状态是有后继状态的,所以说我们只需要记录这两个状态的个数就行了。(所以题目中的概率才用b和d表示么= =)
用dp[id][n][m][a][c]记录id还个人在总高度为n,最上一层有m个(其实这个和n差不多,可以互相推,这样写方便嘛= =),下面的层中A状态有a个,C状态有c个。
然后就记忆化搜索就行了= =
#include <cstdio> #include <cstring> #include <cstdlib> #include <ctime> #include <climits> #include <cmath> #include <iostream> #include <string> #include <vector> #include <set> #include <map> #include <list> #include <queue> #include <stack> #include <deque> #include <algorithm> using namespace std; typedef long long ll; const int maxa = 20; const int maxc = 40; const int maxn = 60; double dp[2][maxn][3][maxa][maxc]; double b[2][3], d[2][3]; inline double P(int i, int j, int n) { return max(0.0, min(1.0, b[i][j] - n * d[i][j])); } double f(int id, int n, int m, int a, int c) { double &ret = dp[id][n][m][a][c]; if (ret >= 0) return ret; if (a == 0 && c == 0) return ret = 0; int n0 = n, a0 = a; if (m == 0) n++; if (m == 2) a++; m = (m + 1) % 3; ret = 0; if (a0 > 0) { ret = max(ret, P(id, 0, n0) * (1.0 - f(1-id, n, m, a-1, c))); ret = max(ret, P(id, 1, n0) * (1.0 - f(1-id, n, m, a-1, c+1))); } if (c > 0) { ret = max(ret, P(id, 2, n0) * (1.0 - f(1-id, n, m, a, c-1))); } return ret; } int main() { int T, n; scanf("%d", &T); while (T--) { for (int i=0;i<2;i++) for (int n=0;n<maxn;n++) for (int m=0;m<3;m++) for (int a=0;a<maxa;a++) for (int c=0;c<maxc;c++) dp[i][n][m][a][c] = -1; scanf("%d", &n); for (int i=0;i<2;i++) for (int j=0;j<3;j++) scanf("%lf%lf", &b[i][j], &d[i][j]); printf("%.4lf\n", f(0, n, 0, n-1, 0)); } return 0; }