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.
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
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
Recommend
哎，思路想复杂了，其实蛮简单的：
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 }