Journey

来源：互联网发布：淘宝企业店铺过户编辑：程序博客网时间：2024/05/22 08:10

http://codeforces.com/contest/839/problem/C

题目大意：

无向图 n个点 n-1条边类似与树结构从1开始到各叶子节点的期望长度从一个点到其相邻节点的概率相同

分析：

dfs求到达叶子节点用路长乘以概率就行

AC代码：

#include <bits/stdc++.h>  #define gcd(a,b) __gcd(a,b)  #define mset(a,x) memset(a,x,sizeof(a))  #define FIN     freopen("jogging.in","r",stdin)  #define FOUT    freopen("jogging.out","w",stdout)  const int INF=0x3f3f3f3f;  const double PI=acos(-1.0);  const int MAX=1e5+10;  typedef long long LL;  const LL mod=1e9+7;  using namespace std;  int cnt;int vis[100005];  int head[100005];int inedge[100005];double ans;void init(){ ans=cnt=0;mset(head,-1);mset(inedge,0);mset(vis,0);}struct edge{int next;int to;}edge[200005];void add(int u,int v){edge[cnt].to=v;edge[cnt].next=head[u];head[u]=cnt++;}void dfs(int u,int countx,double p){int sum=0;for (int i=head[u];i!=-1;i=edge[i].next)if (!vis[edge[i].to]) sum++;if (!sum){ans+=countx*p;return;}for (int i=head[u];i!=-1;i=edge[i].next){if (!vis[edge[i].to]){vis[edge[i].to]=1;dfs(edge[i].to,countx+1,p/sum);}}}int main(){int n;while (scanf ("%d",&n)!=EOF){init();int x,y;for (int i=1;i<n;i++){scanf ("%d%d",&x,&y);add(x,y);add(y,x);inedge[y]=1;}vis[1]=1;dfs(1,0,1);printf ("%.15lf\n",ans);}return 0; }

阅读全文

0 0