成都区域赛区Maze杭电4035(dp求期望)
2012-08-13 20:53
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解题思路分析过程转载 :
02. dp求期望的题。
03. 题意:
04. 有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树,
05. 从结点1出发,开始走,在每个结点i都有3种可能:
06. 1.被杀死,回到结点1处(概率为ki)
07. 2.找到出口,走出迷宫 (概率为ei)
08. 3.和该点相连有m条边,随机走一条
09. 求:走出迷宫所要走的边数的期望值。
10.
11. 设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望。E[1]即为所求。
12.
13. 叶子结点:
14. E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);
15. = ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei);
16.
17. 非叶子结点:(m为与结点相连的边数)
18. E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) );
19. = ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei);
20.
21. 设对每个结点:E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;
22.
23. 对于非叶子结点i,设j为i的孩子结点,则
24. ∑(E[child[i]]) = ∑E[j]
25. = ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
26. = ∑(Aj*E[1] + Bj*E[i] + Cj)
27. 带入上面的式子得
28. (1 - (1-ki-ei)/m*∑Bj)*E[i] = (ki+(1-ki-ei)/m*∑Aj)*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑Cj;
29. 由此可得
30. Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj);
31. Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj);
32. Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj);
33.
34. 对于叶子结点
35. Ai = ki;
36. Bi = 1 - ki - ei;
37. Ci = 1 - ki - ei;
38.
39. 从叶子结点开始,直到算出 A1,B1,C1;
40.
41. E[1] = A1*E[1] + B1*0 + C1;
42. 所以
43. E[1] = C1 / (1 - A1);
44. 若 A1趋近于1则无解...
45.**/
02. dp求期望的题。
03. 题意:
04. 有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树,
05. 从结点1出发,开始走,在每个结点i都有3种可能:
06. 1.被杀死,回到结点1处(概率为ki)
07. 2.找到出口,走出迷宫 (概率为ei)
08. 3.和该点相连有m条边,随机走一条
09. 求:走出迷宫所要走的边数的期望值。
10.
11. 设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望。E[1]即为所求。
12.
13. 叶子结点:
14. E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);
15. = ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei);
16.
17. 非叶子结点:(m为与结点相连的边数)
18. E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) );
19. = ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei);
20.
21. 设对每个结点:E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;
22.
23. 对于非叶子结点i,设j为i的孩子结点,则
24. ∑(E[child[i]]) = ∑E[j]
25. = ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
26. = ∑(Aj*E[1] + Bj*E[i] + Cj)
27. 带入上面的式子得
28. (1 - (1-ki-ei)/m*∑Bj)*E[i] = (ki+(1-ki-ei)/m*∑Aj)*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑Cj;
29. 由此可得
30. Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj);
31. Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj);
32. Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj);
33.
34. 对于叶子结点
35. Ai = ki;
36. Bi = 1 - ki - ei;
37. Ci = 1 - ki - ei;
38.
39. 从叶子结点开始,直到算出 A1,B1,C1;
40.
41. E[1] = A1*E[1] + B1*0 + C1;
42. 所以
43. E[1] = C1 / (1 - A1);
44. 若 A1趋近于1则无解...
45.**/
#include <cstdio> #include <iostream> #include <vector> #include <cmath> using namespace std; const int MAXN = 10000 + 5; double e[MAXN], k[MAXN]; double A[MAXN], B[MAXN], C[MAXN]; vector<int> v[MAXN]; bool search(int i, int fa) { if ( v[i].size() == 1 && fa != -1 ) { A[i] = k[i]; B[i] = 1 - k[i] - e[i]; C[i] = 1 - k[i] - e[i]; return true; } A[i] = k[i]; B[i] = (1 - k[i] - e[i]) / v[i].size(); C[i] = 1 - k[i] - e[i]; double tmp = 0; for (int j = 0; j < (int)v[i].size(); j++) { if ( v[i][j] == fa ) continue; if ( !search(v[i][j], i) ) return false; A[i] += A[v[i][j]] * B[i]; C[i] += C[v[i][j]] * B[i]; tmp += B[v[i][j]] * B[i]; } if ( fabs(tmp - 1) < 1e-10 ) return false; A[i] /= 1 - tmp; B[i] /= 1 - tmp; C[i] /= 1 - tmp; return true; } int main() { int nc, n, s, t; int i; cin >> nc; for (int ca = 1; ca <= nc; ca++) { cin >> n; for (i = 1; i <= n; i++) v[i].clear(); for ( i = 1; i < n; i++) { cin >> s >> t; v[s].push_back(t); v[t].push_back(s); } for ( i = 1; i <= n; i++) { cin >> k[i] >> e[i]; k[i] /= 100.0; e[i] /= 100.0; } cout << "Case " << ca << ": "; if ( search(1, -1) && fabs(1 - A[1]) > 1e-10 ) cout << C[1]/(1 - A[1]) << endl; else cout << "impossible" << endl; } return 0; }
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