Average distance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1765    Accepted Submission(s): 647
Special Judge
 

Problem Description

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 (d01 + d02+ d03 + d04 + d12 +d13 +d14 +d23 +d24 +d34)/10 = (6+3+7+9+9+13+15+10+12+2)/10 = 8.6.

 

 

 

Input

On 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. 
 

 

 

Output

For 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

 

这是一道很简单、经典的题目

让你求出所有点路径之间的距离平均值。

我们可以找任意一条边（红色的边） ，开始算这条边的贡献，可以得到在这条边的左边有3个点，右边4个，那这条边就再结果中做出了 3 * 4 * val(2,5) 的贡献 ，其他边也可以用类似的方法算出贡献来 ，dfs得到一边点的集合个数即可

 

代码如下：

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll,ll>  P;
vector<vector<P> >v;
const int maxn = 2e6;
ll ans[maxn];
int n;
ll val[maxn];
ll sum[maxn],vis[maxn];
int dfs(ll x,ll fa) {
    ll  s = 0;
//    cout << x << endl;
    for(auto const dd:v[x]) {
        ll d = dd.first;
        ll c = dd.second;
        if(d != fa) {
            dfs(d,x);
            vis[x] += vis[d];
            sum[x] += vis[d] * c * (n - vis[d]);
        }
    }

}
int main() {
    int t;
    cin >> t;
    while(t--) {
        cin >> n ;
        v.clear();
        v.resize(n + 500);
        memset(sum,0,sizeof(sum));
        for(int i = 0; i <= n; i++) vis[i] = 1;
        for(int i = 1; i < n; i++) {
            int u,vv,c;
            cin >> u >> vv >> c;
            v[vv].push_back(P(u,c));
            v[u].push_back(P(vv,c));
        }
        dfs(1,-1);
        ll anss = 0;
        for(int i = 0; i <= n; i++) anss += sum[i];
        double ans = 0;
//        cout << anss << endl;
        ans = anss / (double)(n * (n - 1) / 2); 
        printf("%.6f\n",ans);
    }
    return 0;
}


//9
//1 1 2 4 5 6 7 5
//1 -1 -1 3 -1 5 -1 7 7
// 9