HDU 2376 Average distance (树形dp)

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Average distance

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 588 Accepted Submission(s): 213

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

150 1 60 2 30 3 73 4 2

Sample Output

8.6

Source

bapc2007

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2376

题目大意：一棵树，求任意两点间的平均距离

题目分析：先求出任意两点间的距离和，再除总边数(这里的边指的是任意两点间的直接路径数 n * (n - 1) / 2)，求和显然没能枚举，我们可以发现一条路径被经过的次数为其左边点数＊右边点数，所以根据这个我们可以算出每条边对总值的贡献，树形dp求一下即可

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int const MAX = 1e4 + 5;int n, cnt, head[MAX], num[MAX];double sum;struct EDGE{int v, w, next;}e[2 * MAX];void Add(int u, int v, int w){e[cnt].v = v;e[cnt].w = w;e[cnt].next = head[u];head[u] = cnt ++;}void DFS(int u, int fa){num[u] = 1;for(int i = head[u]; i != -1; i = e[i].next){int v = e[i].v;int w = e[i].w;if(v != fa){DFS(v, u);num[u] += num[v];sum += 1.0 * num[v] * (n - num[v]) * w;}}}int main(){int T;scanf("%d", &T);while(T --){cnt = 0;sum = 0;memset(head, -1, sizeof(head));scanf("%d", &n);for(int i = 0; i < n - 1; i++){int u, v, w;scanf("%d %d %d", &u, &v, &w);Add(u, v, w);Add(v, u, w);}DFS(0, -1);printf("%.7f\n", sum / (1.0 * n * (n - 1) / 2.0));}}

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