和ZOJ3329有些像，都是用期望列出来式子以后，为了解式子，设A[i],B[i]，此题又多了C[i]，然后用递推（此题是树形dp）去求得ABC，最后结果只跟ABC有关，跟列写的期望数组根本无关。
虽然式子很长很冗，但平心而论思维上并不难理解，关键是自信和耐心去带入。ABC的递推式出来了以后，代码就不难了。
据说eps有坑？
邝斌巨巨的：
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cmath> 6 #include <ctime> 7 #include <cctype> 8 #include <climits> 9 #include <iostream> 10 #include <iomanip> 11 #include <algorithm> 12 #include <string> 13 #include <sstream> 14 #include <stack> 15 #include <queue> 16 #include <set> 17 #include <map> 18 #include <vector> 19 #include <list> 20 #include <fstream> 21 #include <bitset> 22 #define init(a, b) memset(a, b, sizeof(a)) 23 #define rep(i, a, b) for (int i = a; i <= b; i++) 24 #define irep(i, a, b) for (int i = a; i >= b; i--) 25 using namespace std; 26 27 typedef double db; 28 typedef long long ll; 29 typedef unsigned long long ull; 30 typedef pair<int, int> P; 31 const int inf = 0x3f3f3f3f; 32 const ll INF = 1e18; 33 34 template <typename T> void read(T &x) { 35 x = 0; 36 int s = 1, c = getchar(); 37 for (; !isdigit(c); c = getchar()) 38 if (c == '-') s = -1; 39 for (; isdigit(c); c = getchar()) 40 x = x * 10 + c - 48; 41 x *= s; 42 } 43 44 template <typename T> void write(T x) { 45 if (x < 0) x = -x, putchar('-'); 46 if (x > 9) write(x / 10); 47 putchar(x % 10 + '0'); 48 } 49 50 template <typename T> void writeln(T x) { 51 write(x); 52 puts(""); 53 } 54 55 const int maxn = 1e4 + 5; 56 const db eps = 1e-9; 57 int T, n, kase; 58 db A[maxn], B[maxn], C[maxn], K[maxn], E[maxn]; 59 vector<int> v[maxn]; 60 61 bool dfs(int cur, int fa) { 62 int m = v[cur].size(); 63 if (m == 1 && fa) { 64 A[cur] = K[cur]; 65 B[cur] = C[cur] = 1.0 - K[cur] - E[cur]; 66 } else { 67 db tmp = (1 - K[cur] - E[cur]) / m; 68 db Atmp = 0, Btmp = 0, Ctmp = 0; 69 for (int child : v[cur]) { 70 if (child != fa) { 71 if (!dfs(child, cur)) return false; 72 Atmp += A[child]; 73 Btmp += B[child]; 74 Ctmp += C[child]; 75 } 76 } 77 if (fabs(1 - tmp * Btmp) < eps) return false; 78 A[cur] = (K[cur] + tmp * Atmp) / (1 - tmp * Btmp); 79 B[cur] = tmp / (1 - tmp * Btmp); 80 C[cur] = (tmp * Ctmp + 1 - K[cur] - E[cur]) / (1 - tmp * Btmp); 81 } 82 return true; 83 } 84 85 int main() { 86 for (read(T); T; T--) { 87 read(n); 88 rep(i, 1, n) v[i].clear(); 89 rep(i, 1, n - 1) { 90 int x, y; 91 read(x), read(y); 92 v[x].push_back(y); 93 v[y].push_back(x); 94 } 95 rep(i, 1, n) { 96 read(K[i]), read(E[i]); 97 K[i] /= 100, E[i] /= 100; 98 } 99 100 printf("Case %d: ", ++kase); 101 if (dfs(1, 0) && fabs(1 - A[1]) > eps) { 102 printf("%.6lf\n", C[1] / (1 - A[1])); 103 } else { 104 puts("impossible"); 105 } 106 } 107 return 0; 108 }
