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Computer

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

5 1 1 2 1 3 1 1 1

 

Sample Output

3 2 3 4 4

 

---

　　分析：

　　一道经典的树形DP。

　　每次从需要求的节点出发，然后遍历该树，期间更新该点的最长路。但是用DP数组存最长路的时候有一个问题，就是便利的方向，如果不考虑方向的话肯定是错的，那么可以转换一下，DP数组不存点的最长路，存边的最长路，这样使用邻接链表村边的化就可以考虑到方向了。具体看代码。

　　Code：

//It is made by HolseLee on 24th July 2018
//HDU2196
#include<bits/stdc++.h>
using namespace std;

const int N=1e4+7;
int n,dp[N<<1],head[N],size;
struct Node{
    int to,next,val;
}edge[N<<1];

inline int read()
{
    char ch=getchar();int num=0;bool flag=false;
    while(ch<'0'||ch>'9'){if(ch=='-')flag=true;ch=getchar();}
    while(ch>='0'&&ch<='9'){num=num*10+ch-'0';ch=getchar();}
    return flag?-num:num;
} 

inline void add(int x,int y,int z)
{
    edge[++size].to=y;
    edge[size].val=z;
    edge[size].next=head[x];
    head[x]=size;
}

inline int dfs(int x,int las)
{
    int ret=0;
    for(int i=head[x];i!=-1;i=edge[i].next){
        int y=edge[i].to;
        if(y==las)continue;
        if(!dp[i]){
            dp[i]=max(dp[i],dfs(y,x)+edge[i].val);
        }
        ret=max(ret,dp[i]);
    }
    return ret;
} 

;int main()
{
    while(scanf("%d",&n)!=EOF){
        int x,y;
        memset(dp,0,sizeof(dp));
        memset(head,-1,sizeof(head));
        size=0;
        for(int i=1;i<n;i++){
            x=read();y=read();
            add(i+1,x,y);add(x,i+1,y);
        }
        for(int i=1;i<=n;i++)
        printf("%d
",dfs(i,i));
    }
}