cf280C. Game on Tree(期望线性性)
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2022-06-30 18:19:08
题意 "题目链接" Sol 开始想的dp,发现根本不能转移(貌似只能做链) 根据期望的线性性,其中$ans = \sum_{1 f(x)}$ $f(x)$表示删除$x$节点的概率,显然$x$节点要被删除,那么它的祖先都不能被删除,因此概率为$\frac{1}{deep[x]}$ cpp includ ......
题意
sol
开始想的dp,发现根本不能转移(貌似只能做链)
根据期望的线性性,其中\(ans = \sum_{1 * f(x)}\)
\(f(x)\)表示删除\(x\)节点的概率,显然\(x\)节点要被删除,那么它的祖先都不能被删除,因此概率为\(\frac{1}{deep[x]}\)
#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; }
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