7.15 K codeforces 696BPuzzle
来源:互联网 发布:淘宝退款几天到银行卡 编辑:程序博客网 时间:2024/06/05 15:02
题意:给一棵有根树,dfs给节点标号,问每个节点的编号期望。
思路:观察发现节点编号期望与子节点个数有关,根据样例可猜得关系表达式。
代码:
#include<algorithm>#include<cstring>#include<cstdio>#include<vector>using namespace std;const int maxn=100001;int p[maxn],s[maxn];double ans[maxn];int n;int main(){ scanf("%d",&n); for(int i=2;i<=n;i++) scanf("%d",&p[i]); for(int i=1;i<=n+1;i++) {s[i]=1;ans[i]=1.0;} for(int i=n;i>=2;i--) s[p[i]]+=s[i]; printf("%lf ",ans[1]=1.0); for(int i=2;i<=n;i++) printf("%lf ",ans[i]=1.0+ans[p[i]]+0.5*(s[p[i]]-s[i]-1)); return 0;}
0 0
- 7.15 K codeforces 696BPuzzle
- Codeforces---k-String
- codeforces 630 K. Indivisibility
- codeforces 630K Indivisibility
- codeforces 730K
- k-th divisor CodeForces
- k-th divisor CodeForces
- codeforces A. k-Factorization
- Codeforces 630K
- codeforces [Gym-100814K]
- K-Dominant Character CodeForces
- K-Dominant Character CodeForces
- codeforces A. k-String 题解
- Codeforces - 219A - k-String
- CodeForces 431C K-Tree
- CodeForces 219A k-String
- codeforces Problem- 630K Indivisibility
- CodeForces 431C k-Tree
- Codeforces Round #370 (Div. 2) C 贪心
- 算法学习
- vijos p1347&&vijos p1455(递推)
- 外销升级接口V2.1.3TPS波动较大-基于索引的sql优化
- Tomcat源码分析(一)--服务启动
- 7.15 K codeforces 696BPuzzle
- [模板]DFS序的应用-初级
- ImportREC输入表重建工具
- geoip 添加一列,add_field =>["[geoip][request_time]","%{request_time}"]
- Tomcat源码分析(二)--连接处理
- Python-json模块
- poj 2828 线段树(单点更新,逆序插入)
- 169\229- Majority Element
- Linux netstat命令详解