hdu4035_概率dp
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4035
dp求期望的题。 题意: 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, 从结点1出发,开始走,在每个结点i都有3种可能: 1.被杀死,回到结点1处(概率为ki) 2.找到出口,走出迷宫 (概率为ei) 3.和该点相连有m条边,随机走一条 求:走出迷宫所要走的边数的期望值。 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。 叶子结点: E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1); = ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei); 非叶子结点:(m为与结点相连的边数) E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) ); = ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei); 设对每个结点:E[i] = Ai*E[1] + Bi*E[father[i]] + Ci; 对于非叶子结点i,设j为i的孩子结点,则 ∑(E[child[i]]) = ∑E[j] = ∑(Aj*E[1] + Bj*E[father[j]] + Cj) = ∑(Aj*E[1] + Bj*E[i] + Cj) 带入上面的式子得 (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; 由此可得 Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj); Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj); Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj); 对于叶子结点 Ai = ki; Bi = 1 - ki - ei; Ci = 1 - ki - ei; 从叶子结点开始,直到算出 A1,B1,C1; E[1] = A1*E[1] + B1*0 + C1; 所以 E[1] = C1 / (1 - A1); 若 A1趋近于1则无解...
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cstdlib> 6 #include <cmath> 7 #include <set> 8 #include <map> 9 #include <queue> 10 #include <vector> 11 using namespace std; 12 13 const int N = 10000 + 10; 14 double e[N], k[N], a[N], b[N], c[N]; 15 vector <int> v[N]; 16 bool find(int i, int fa) 17 { 18 if(v[i].size() == 1 && fa != -1) 19 { 20 a[i] = k[i]; 21 b[i] = 1 - k[i] - e[i]; 22 c[i] = 1 - k[i] - e[i]; 23 return true; 24 } 25 a[i] = k[i]; 26 b[i] = (1 - k[i] - e[i]) / v[i].size(); 27 c[i] = 1 - k[i] - e[i]; 28 double temp = 0; 29 for(int j = 0; j < (int)v[i].size(); j++) 30 { 31 if(v[i][j] == fa) 32 continue; 33 if(!find(v[i][j], i)) 34 return false; 35 a[i] += b[i] * a[v[i][j]]; 36 c[i] += b[i] * c[v[i][j]]; 37 temp += b[i] * b[v[i][j]]; 38 } 39 if(fabs(1 - temp) < 1e-10) 40 return false; 41 a[i] /= (1 - temp); 42 b[i] /= (1 - temp); 43 c[i] /= (1 - temp); 44 return true; 45 } 46 int main() 47 { 48 int t, n, x, y, cnt = 0; 49 scanf("%d", &t); 50 while(t--) 51 { 52 cnt++; 53 scanf("%d", &n); 54 for(int i = 1; i <= n; i++) 55 v[i].clear(); 56 for(int i = 1; i < n; i++) 57 { 58 scanf("%d %d", &x, &y); 59 v[x].push_back(y); 60 v[y].push_back(x); 61 } 62 for(int i = 1; i <= n; i++){ 63 scanf("%lf %lf", k + i, e + i); 64 k[i] /= 100.0; 65 e[i] /= 100.0; 66 } 67 cout << "Case " << cnt << ": "<<endl; 68 if(find(1, -1) && 1 - a[1] > 1e-10) 69 printf("%.6lf\n", c[1]/(1 - a[1])); 70 else 71 cout << "impossible" << endl; 72 } 73 return 0; 74 }