hdu2196 Computer(求树的直径)

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3583    Accepted Submission(s): 1818

Problem Description

A school bought the first computer some time ago(so this computer's id is 1). During the recent years the school bought N-1 new computers. Each new computer was connected to one of settled earlier. Managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which i-th computer needs to send signal (i.e. length of cable to the most distant computer). You need to provide this information. 


Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.

 

Input

Input file contains multiple test cases.In each case there is natural number N (N<=10000) in the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two natural numbers - number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input are separated by a space.

 

Output

For each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N).

 

Sample Input

51 12 13 11 1

 

Sample Output

32344

 

题意：一个学校采购了N台电脑，每两台电脑直接连接，各电脑分别从1到N.两两距离为v，问每台电脑到这些电脑哪台是最远距离为多少？

题解：记得以前和LSS做过一个关于求树的直径的问题，这个题目类似，先求出数的直径的两个端点，分别求出这两个端点到其他点的距离，其他点最大的距离一定是这两个端点二选一。用3个深搜就可以了。

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <queue>#include <map>#include <stack>#include <list>#include <vector>#include <ctime>#define LL __int64#define eps 1e-8using namespace std;struct node{int a,b,v,next;}edge[20005];int head[20005],w,dp[20005],ans[20005];void add(int a,int b,int v){w++;edge[w].a=a;edge[w].b=b;edge[w].v=v;edge[w].next=head[a];head[a]=w;}void DFS(int x,int father){for (int i=head[x];i!=-1;i=edge[i].next){int to=edge[i].b;if (to==father) continue;dp[to]=max(dp[to],dp[x]+edge[i].v);DFS(to,x);}}int main(){int n,i,j,v;while (~scanf("%d",&n)){w=0;memset(dp,0,sizeof(dp));memset(head,-1,sizeof(head));memset(edge,0,sizeof(edge));for (i=2;i<=n;i++){scanf("%d%d",&j,&v);add(i,j,v);add(j,i,v);}DFS(1,-1);int k=-1,l1,l2;for (i=1;i<=n;i++){if (k<dp[i]){k=dp[i];l1=i;}dp[i]=0;}DFS(l1,-1);k=-1;for (i=1;i<=n;i++){if (k<dp[i]){l2=i;k=dp[i];}ans[i]=dp[i];dp[i]=0;}DFS(l2,-1);for (i=1;i<=n;i++)printf("%d\n",max(ans[i],dp[i]));}return 0;}