期望DP

本題遞推比較麻煩,可以記憶化搜索
注意搜索的邊界條件
以及每一次轉移

#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 ,所以我們直接加 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; }

轉載于:https://www.cnblogs.com/Mr-WolframsMgcBox/p/8651281.html

總結

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