Description
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。
Solution
这题其实是个套路。
考虑期望要倒推,我们设
表示已经走到
了,剩下步数的期望。
我们先套路地设
设
,且下面式子中的
表示的均为
的子节点,根据题意有:
将 带入 得:
化简发现:
有点丑不好意思
对于根,我们有:
解得最后的答案为
Code
/**************************************
* Au: Hany01
* Prob: [HDU4035] Maze
* Date: Jun 9th, 2018
* Email: [email protected]
**************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = j; i < i##_end_; ++ i)
#define For(i, j ,k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define y1 wozenmezhemecaia
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
inline int read() {
register char c_; register int _, __;
for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}
const int maxn = 10005, maxm = 20005;
const double eps = 1e-9;
int beg[maxn], nex[maxm], e, v[maxm], d[maxn];
double A[maxn], B[maxn], C[maxn], p[maxn], k[maxn];
inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e, ++ d[vv]; }
void DFS(int u, int pa)
{
double totb = 0, totc = 0, tota = 0;
for (register int i = beg[u]; i; i = nex[i]) if (v[i] ^ pa)
DFS(v[i], u), tota += A[v[i]], totb += B[v[i]], totc += C[v[i]];
A[u] = p[u] / d[u], B[u] = k[u] + p[u] * totb / d[u], C[u] = p[u] + p[u] * totc / d[u];
tota = 1 - p[u] * tota / d[u];
A[u] /= tota, B[u] /= tota, C[u] /= tota;
}
int main()
{
#ifdef hany01
File("hdu4035");
#endif
static int uu, vv, n;
for (static int T = read(), Case = 1; Case <= T; ++ Case) {
n = read(), Set(beg, 0), Set(d, 0), e = 0;
For(i, 2, n) uu = read(), vv = read(), add(uu, vv), add(vv, uu);
For(i, 1, n) k[i] = read() / 100., p[i] = 1 - k[i] - read() / 100.;
DFS(1, 0);
if (fabs(1 - B[1]) < eps) printf("Case %d: impossible\n", Case);
else printf("Case %d: %.6lf\n", Case, C[1] / (1 - B[1]));
}
return 0;
}