hdu 2736 Average distance

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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lcy | We have carefully selected several similar problems for you: 2378 2379 2377 2380 2381

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

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <stack>
 6 #include <cctype>
 7 #include <vector>
 8 #include <cmath>
 9 #include <map>
10 #include <queue>
11 
12 #define ll long long
13 #define N 10005
14 #define eps 1e-8
15 
16 using namespace std;
17 
18 int T;
19 double tot[N];
20 double cou[N];
21 double num;
22 int n;
23 
24 struct PP
25 {
26     int to;
27     double val;
28 };
29 
30 vector<PP>Edge[N];
31 
32 PP te;
33 
34 void add_adge(int from,int to,double val)
35 {
36     te.to = to;te.val = val;
37     Edge[from].push_back(te);
38     te.to = from;te.val = val;
39     Edge[to].push_back(te);
40 }
41 
42 void dfs(int now,int fa)
43 {
44     unsigned int i;
45     PP nt;
46     cou[now] = 1;
47     for(i = 0;i < Edge[now].size();i++){
48         nt.to = Edge[now][i].to;
49         nt.val = Edge[now][i].val;
50         if(nt.to == fa){
51             continue;
52         }
53         dfs(nt.to,now);
54         tot[now] = tot[now] + tot[nt.to] + (n-cou[nt.to]) * cou[nt.to] * nt.val;
55         cou[now] = cou[now] + cou[nt.to];
56         //printf("  now = %d i=%d to=%d tot=%.6f cou=%.6lf
",now,i,nt.to,tot[now],cou[now]);
57     }
58     //printf(" i=%d tot=%.6f cou=%.6f
",now,tot[now],cou[now]);
59 }
60 
61 int main()
62 {
63     //freopen("in.txt","r",stdin);
64     scanf("%d",&T);
65     int i;
66     int from,to;
67     double val;
68     for(int ccnt=1;ccnt<=T;ccnt++){
69     //while(scanf("%lf%lf%lf%lf",&a[0],&a[1],&a[2],&a[3])!=EOF){
70         scanf("%d",&n);
71         memset(tot,0,sizeof(tot));
72         memset(cou,0,sizeof(cou));
73         num = 1.0 *n*(n-1)/2;
74         for(i=0;i<=n;i++){
75             Edge[i].clear();
76         }
77         for(i=1;i<=n-1;i++){
78             scanf("%d%d%lf",&from,&to,&val);
79             add_adge(from,to,val);
80         }
81         dfs(0,-1);
82         //for(i=0;i<n;i++){
83          //   for(int j=0;j<Edge[i].size();j++){
84          //       printf(" i=%d to=%d val=%.6lf
",i,Edge[i][j].to,Edge[i][j].val);
85          //   }
86        // }
87         for(i=0;i<n;i++){
88             //printf(" i=%d tot=%.6f cou=%.6f
",i,tot[i],cou[i]);
89         }
90         printf("%lf
",tot[0]/num);
91     }
92     return 0;
93 }