cf280C. Game on Tree(期望線性性)
阿新 • • 發佈:2019-01-08
題意
Sol
開始想的dp,發現根本不能轉移(貌似只能做鏈)
根據期望的線性性,其中\(ans = \sum_{1 * f(x)}\)
\(f(x)\)表示刪除\(x\)節點的概率,顯然\(x\)節點要被刪除,那麼它的祖先都不能被刪除,因此概率為\(\frac{1}{deep[x]}\)
#include<bits/stdc++.h> #define Pair pair<int, int> #define MP(x, y) make_pair(x, y) #define fi first #define se second //#define int long long #define LL long long #define ull unsigned long long #define Fin(x) {freopen(#x".in","r",stdin);} #define Fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10; const double eps = 1e-9; template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;} template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;} template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;} template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;} template <typename A> inline void debug(A a){cout << a << '\n';} template <typename A> inline LL sqr(A x){return 1ll * x * x;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, dep[MAXN]; vector<int> v[MAXN]; void dfs(int x, int fa) { dep[x] = dep[fa] + 1; for(int i = 0; i < v[x].size(); i++) { int to = v[x][i]; if(to == fa) continue; dfs(to, x); } } signed main() { N = read(); for(int i = 1; i <= N - 1; i++) { int x = read(), y = read(); v[x].push_back(y); v[y].push_back(x); } dfs(1, 0); double ans = 0; for(int i = 1; i <= N; i++) ans += 1.0 / dep[i]; printf("%.12lf", ans); return 0; }