Average distance(类树形DP)

Average distance

HDU - 2376

Given a tree, calculate the average distance between two vertices in the tree. For example, the average distance between two vertices in the following tree is (d ₀₁ + d ₀₂ + d ₀₃ + d ₀₄ + d ₁₂ +d ₁₃ +d ₁₄ +d ₂₃ +d ₂₄ +d ₃₄)/10 = (6+3+7+9+9+13+15+10+12+2)/10 = 8.6.

InputOn the first line an integer t (1 <= t <= 100): the number of test cases. Then for each test case:

One line with an integer n (2 <= n <= 10 000): the number of nodes in the tree. The nodes are numbered from 0 to n - 1.

n - 1 lines, each with three integers a (0 <= a < n), b (0 <= b < n) and d (1 <= d <= 1 000). There is an edge between the nodes with numbers a and b of length d. The resulting graph will be a tree.

OutputFor each testcase:

One line with the average distance between two vertices. This value should have either an absolute or a relative error of at most 10 ^-6

Sample Input

1
5
0 1 6
0 2 3
0 3 7
3 4 2

Sample Output

8.6题意：求一个树上任意两点距离的平均值。其实也不算是一棵树，因为节点之间都是双向边。思路：　　对于每一条边，计算一下它两端的节点数A,B,那么每一条边的权值*左边的节点数*右边的节点数就是该条边的被经过的次数累加的贡献。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<vector>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef long long ll;
 8 const int maxn=1e5+10;
 9 struct node{
10     int v;
11     ll w;
12 }now;
13 vector<node>v[maxn];
14 int vis[maxn];
15 int n,a,b;
16 ll w,ans,t,son[maxn];
17 void dfs(int x)
18 {
19     vis[x]=1;
20     son[x]=1;
21     for(int i=0;i<v[x].size();i++)
22     {
23         node aa=v[x][i];
24         ll len=aa.w;
25         int to=aa.v;
26         if(vis[to])
27             continue;
28         dfs(to);
29         son[x]+=son[to];
30         ans+=len*(n-son[to])*son[to]; //注意2：这里一定要写son[to]而不是son[x] 写son[x]会导致边和其两边的点数不匹配。
31     //    cout<<x<<" "<<to<<endl;
32         cout<<ans<<endl;
33     }
34 }
35 int main()
36 {
37     int casen;
38     cin>>casen;
39     while(casen--)
40     {
41         ans=0;
42         memset(son,0,sizeof(son));
43         memset(vis,0,sizeof(vis));
44         scanf("%d",&n);
45         for(int i=0;i<=n;i++)
46             v[i].clear();
47         for(int i=1;i<n;i++)
48         {
49             scanf("%d%d%lld",&a,&b,&w);
50             now.v=b;
51             now.w=w;
52             v[a].push_back(now);
53             now.v=a;                //注意1：双向边对于该边的两个端点都要标记权值为边的值，这样不管先遍历到哪个节点，这个节点的权值就是该点和其下一个点之间的边的权值。
54             v[b].push_back(now);
55         }
56         dfs(0);
57         t=(n-1)*n/2;
58         printf("%.11f\n",(double)ans/(double)t);//注意3： 题目好坑啊，要求的误差在10-6 但是样例给的只是小数点后一位。
59     }
60 }

原文地址：https://www.cnblogs.com/1013star/p/9959873.html