Average distance(类树形DP)
Average distance
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 }
