hdu 2736 Average distance

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Average distance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 682    Accepted Submission(s): 244
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

 

Source

bapc2007

 

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哎，思路想复杂了，其实蛮简单的：

转一发题解：

引：如果暴力枚举两点再求距离是显然会超时的。转换一下思路，我们可以对每条边，求所有可能的路径经过此边的次数：设这条边两端的点数分别为A和B，那 么这条边被经过的次数就是A*B，它对总的距离和的贡献就是(A*B*此边长度)。我们把所有边的贡献求总和，再除以总路径数N*(N-1)/2，即为最 后所求。

每条边两端的点数的计算，实际上是可以用一次dfs解决的。任取一点为根，在dfs的过程中，对每个点k记录其子树包含的点数（包括其自身），设点数为a[k]，则k的父亲一侧的点数即为N-a[k]。这个统计可以和遍历同时进行。故时间复杂度为O(n)。

16447376

2016-03-06 09:40:59

Accepted

2376

312MS

4540K

2093 B

C++

czy

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <stack>
 #include <cctype>
 #include <vector>
 #include <cmath>
 #include <map>
 #include <queue>

 #define ll long long
 #define N 10005
 #define eps 1e-8

 using namespace std;

 int T;
 double tot[N];
 double cou[N];
 double num;
 int n;

 struct PP
 {
     int to;
     double val;
 };

 vector<PP>Edge[N];

 PP te;

 void add_adge(int from,int to,double val)
 {
     te.to = to;te.val = val;
     Edge[from].push_back(te);
     te.to = from;te.val = val;
     Edge[to].push_back(te);
 }

 void dfs(int now,int fa)
 {
     unsigned int i;
     PP nt;
     cou[now] = ;
     for(i = ;i < Edge[now].size();i++){
         nt.to = Edge[now][i].to;
         nt.val = Edge[now][i].val;
         if(nt.to == fa){
             continue;
         }
         dfs(nt.to,now);
         tot[now] = tot[now] + tot[nt.to] + (n-cou[nt.to]) * cou[nt.to] * nt.val;
         cou[now] = cou[now] + cou[nt.to];
         //printf("  now = %d i=%d to=%d tot=%.6f cou=%.6lf\n",now,i,nt.to,tot[now],cou[now]);
     }
     //printf(" i=%d tot=%.6f cou=%.6f\n",now,tot[now],cou[now]);
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     scanf("%d",&T);
     int i;
     int from,to;
     double val;
     for(int ccnt=;ccnt<=T;ccnt++){
     //while(scanf("%lf%lf%lf%lf",&a[0],&a[1],&a[2],&a[3])!=EOF){
         scanf("%d",&n);
         memset(tot,,sizeof(tot));
         memset(cou,,sizeof(cou));
         num = 1.0 *n*(n-)/;
         for(i=;i<=n;i++){
             Edge[i].clear();
         }
         for(i=;i<=n-;i++){
             scanf("%d%d%lf",&from,&to,&val);
             add_adge(from,to,val);
         }
         dfs(,-);
         //for(i=0;i<n;i++){
          //   for(int j=0;j<Edge[i].size();j++){
          //       printf(" i=%d to=%d val=%.6lf\n",i,Edge[i][j].to,Edge[i][j].val);
          //   }
        // }
         for(i=;i<n;i++){
             //printf(" i=%d tot=%.6f cou=%.6f\n",i,tot[i],cou[i]);
         }
         printf("%lf\n",tot[]/num);
     }
     return ;
 }