【题目链接】:

【题意】

让你计算树上任意两点之间的距离的和.

【题解】

算出每条边的两端有多少个节点设为num1和num2;

这条边的边权为w;

答案累加上w*num1*num2;

然后总的答案除n*(n-1)/2;

【完整代码】

#include 
     
      using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%lld",&x)#define ref(x) scanf("%lf",&x)typedef pair
      
        pii;typedef pair
       
         pll;const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };const double pi = acos(-1.0);const int N = 1e4+100;int n;LL sum[N];double ans;vector 
        
          G[N];vector 
         
           w[N];void in(){    rei(n);    rep1(i, 1, n)        G[i].clear(), w[i].clear();    rep1(i, 1, n - 1)    {        int x, y; LL z;        rei(x), rei(y), rel(z);        x++, y++;        G[x].push_back(y),G[y].push_back(x);        w[x].push_back(z), w[y].push_back(z);    }}void dfs(int x, int fa){    sum[x] = 1;    int len = G[x].size();    rep1(i,0,len-1)    {        int y = G[x][i];        if (y == fa) continue;        dfs(y, x);        sum[x] += sum[y];        ans += 1LL*(n - sum[y])*sum[y] * w[x][i];    }}void o(){    double tt = n*(n - 1) / 2;    ans /= tt;    printf("%.8f\n", ans);}int main(){    //freopen("F:\\rush.txt", "r", stdin);    int t;    rei(t);    while (t--)    {        ans = 0;        in();        dfs(1, 0);        o();    }    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);    return 0;}