HDU-2196-Computer

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31489    Accepted Submission(s): 4111

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：求各个点能够到达的最远距离。

  节点能到达的最远距离考虑往上父节点方向，往下儿子节点方向，然后搜两次，第一次求从父节点方向的最远距离，第二次儿子节点方向的最远距离。这题也算是入门题吧，不过当时懵懵懂懂刚开始学，用的word中的写法。

代码：

  

#include<iostream>#include<string.h>#include<math.h>#include<stdio.h>#include<vector>using namespace std;vector<int>son[10005],v[10005];int bj[10005],g[10005],ll[10005],h[10005];int dfs(int p){int flag;if(bj[p])    return bj[p];int l=son[p].size();if(l==0)    return 0;for(int i=0;i<l;i++)    {    int j=son[p][i];    if(dfs(j)+v[p][i]>bj[p])        {        g[p]=bj[p];        bj[p]=bj[j]+v[p][i];        flag=j;        }    else  if(bj[j]+v[p][i]>g[p])            g[p]=bj[j]+v[p][i];    }ll[p]=flag;return bj[p];}void dfs1(int p){int l=son[p].size();for(int i=0;i<l;i++)    {    int j=son[p][i];    if(j==ll[p])        h[j]=max(h[p],g[p])+v[p][i];    else        h[j]=max(h[p],bj[p])+v[p][i];    dfs1(j);    }}int main(){int n;int a,b;while(cin>>n)    {    memset(g,0,sizeof(g));    memset(h,0,sizeof(h));    memset(bj,0,sizeof(bj));    for(int i=1;i<=n;i++)        {        son[i].clear();        v[i].clear();        }    for(int i=2;i<=n;i++)        {        scanf("%d%d",&a,&b);        son[a].push_back(i);        v[a].push_back(b);        }    dfs(1);    dfs1(1);    for(int i=1;i<=n;i++)        cout<<max(bj[i],h[i])<<endl;    }return 0;}