HDU 2736 Average Distance

Transmission Door

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 Descriptiongiven a tree, calculate the average distance between, vertices in the tree. For example, the average distance between and the vertices in the following tree are (d01 + d02 + d03 + d04 + D12 +d13 +d14 +d2 3 +d24 +d34)/10 = (6+3+7+9+9+13+15+10+12+2)/10 = 8.6.

Inputon the first line an integer t (1 <= t <=): The number of test cases. Then to each test case:

One line with a integer n (2 <= n <=): The number of nodes in the tree. The nodes is 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 would be a tree.

Outputfor each testcase:

One line with the average distance between and the vertices. This value should has either an absolute or a relative error of in most 10-6

Sample Input150 1 60 2 30 3 73 4 2

Sample Output8.6

Source bapc2007

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Ah, thinking of complex, in fact, quite simple:

Turn the puzzle:

Citation: If the brute force enumeration is two points, then the distance is obviously timed out. Switching ideas, we can on each side, to ask all possible paths through this side of the number of times: Set the end of this side of the points are a and B, then the number of times the edge is a*b, it is the total distance and contribution is (a*b* this side length). We sum up the contribution of all sides, divided by the total number of paths N (N-1)/2, which is the last request.

The calculation of the number of points at each end of each edge can actually be resolved with Dfs one time. Whichever is the root, in the DFS process, for each point K records its subtree contains the number of points (including its own), the points is a[k], then K on the father side of the point is n-a[k]. This statistic can be performed concurrently with traversal. Therefore, the time complexity is 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>Ten#include <queue> One  A #definell Long Long - #defineN 10005 - #defineEPS 1e-8 the  - using namespacestd; -  - intT; + DoubleTot[n]; - DoubleCou[n]; + Doublenum; A intN; at  - structPP - { -     intto ; -     DoubleVal; - }; in  -Vector<pp>Edge[n]; to  + PP te; -  the voidAdd_adge (int  from,intTo,Doubleval) * { $te.to = To;te.val =Val;Panax Notoginsengedge[ from].push_back (TE); -Te.to = from; te.val =Val; the edge[to].push_back (TE); + } A  the voidDfsintNowintFA) + { -Unsignedinti; $ PP NT; $Cou[now] =1; -      for(i =0; i < edge[now].size (); i++){ -Nt.to =edge[now][i].to; theNt.val =Edge[now][i].val; -         if(Nt.to = =FA) {Wuyi             Continue; the         } - DFS (nt.to,now); WuTot[now] = Tot[now] + tot[nt.to] + (n-cou[nt.to]) * cou[nt.to] *Nt.val; -Cou[now] = Cou[now] +cou[nt.to]; About         //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]); - } -  A intMain () + { the     //freopen ("In.txt", "R", stdin); -scanf"%d",&T); $     inti; the     int  from, to; the     DoubleVal; the      for(intCcnt=1; ccnt<=t;ccnt++){ the     //while (scanf ("%lf%lf%lf%lf", &a[0],&a[1],&a[2],&a[3])!=eof) { -scanf"%d",&n); inmemset (Tot,0,sizeof(tot)); thememset (Cou,0,sizeof(Cou)); thenum =1.0*n* (n1)/2; About          for(i=0; i<=n;i++){ the edge[i].clear (); the         } the          for(i=1; i<=n-1; i++){ +scanf"%D%D%LF",& from,&to,&val); -Add_adge ( from, to,val); the         }BayiDfs0,-1); the         //For (i=0;i<n;i++) { the          //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); -          //   } the        // } the          for(i=0; i<n;i++){ the             //printf ("i=%d tot=%.6f cou=%.6f\n", I,tot[i],cou[i]); the         } -printf"%lf\n", tot[0]/num); the     } the     return 0; the}

HDU 2736 Average Distance