Computer （树形dp）

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

题目大概：

有一颗有边权的有根树，求树中每个节点和离它最远的点的距离。

思路：

一开始以为只是求这颗树的最大距离，直接根据小白书上的一次dfs，结果发现输出有很多，原来是要求每个节点的最远距离。

要用两次dfs，

第一次dfs，求每个节点在它儿子方向的最大距离，即从该节点往下出发（叶子方向），找最大距离。

这次dfs利用孩子节点的信息跟新本节点的最大距离和次最大距离。

第二次dfs，求每个节点在它父亲方向的最大距离，即从给节点往上走（根的方向），找最大距离。

这次dfs利用本结点跟新孩子节点最大距离和次最大距离。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;int n;int dp[10005][2];int head[10005];int minshu;int ans;struct shu{    int v;    int next;    int ju;}tr[20010];void add(int q,int w,int ju){    tr[ans].v=w;    tr[ans].next=head[q];    tr[ans].ju=ju;    head[q]=ans++;}void inset(int v,int &x0,int &x1){    if(v>x0)    {        x1=x0;        x0=v;    }    else if(v>x1)    {        x1=v;    }}void dfs(int x,int pa){    dp[x][0]=dp[x][1]=0;    for(int i=head[x];i!=-1;i=tr[i].next)    {        int son=tr[i].v;        if(son!=pa)        {            dfs(son,x);            int ww=tr[i].ju;            inset(dp[son][0]+ww,dp[x][0],dp[x][1]);        }    }}void dfs1(int x,int pa){    for(int i=head[x];i!=-1;i=tr[i].next)    {        int son=tr[i].v;        if(son!=pa)        {            int ww=tr[i].ju;            if(dp[son][0]+ww==dp[x][0])            {                 inset(dp[x][1]+ww,dp[son][0],dp[son][1]);            }            else            {                 inset(dp[x][0]+ww,dp[son][0],dp[son][1]);            }            dfs1(son,x);        }    }}int main(){    while(~scanf("%d",&n))    {        memset(dp,0,sizeof(dp));          memset(head,-1,sizeof(head));        ans=0;        for(int i=2;i<=n;i++)        {            int q,w;            scanf("%d%d",&q,&w);            add(q,i,w);            add(i,q,w);        }        dfs(1,-1);        dfs1(1,-1);        for(int i=1;i<=n;i++)    {           printf("%d\n",dp[i][0]);    }    }    return 0;}