hdu 2196 Computer 樹形dp模板題

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2850    Accepted Submission(s): 1450

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：以編號的i的節點爲例（非根節點），最長的路徑長度只有倆種可能，
    1）子樹中存在最長路徑；
    2）通過父節點的路徑中存在最長路徑
所以，只有分別求出每一節點對應的那倆種路徑取大最大值即可，其中，根節點只存在第一種可能
***/
#include "stdio.h"  
#include "string.h"

#define N 10005

struct node{
    int x,y;
    int weight;
    int next;
}edge[4*N];
int idx,head[N];

void Init(){idx=0; memset(head,-1,sizeof(head));}

void Add(int x,int y,int weight)
{
    edge[idx].x = x;
    edge[idx].y = y;
    edge[idx].weight = weight;
    edge[idx].next = head[x];
    head[x] = idx++;
}

struct point{
    int id;
    int value;
}dp1[N],dp2[N];  //dp1[i]記錄點i的最遠距離，dp2[i]記錄點i的次遠距離，

void swap(point &a,point &b)
{
    point c;
    c = a;
    a = b;
    b = c;
}

void DFS1(int x,int father)
{
    int i,y;
    dp1[x].value = dp2[x].value = 0;
    for(i=head[x]; i!=-1; i=edge[i].next)
    {
        y = edge[i].y;
        if(y==father) continue;
        DFS1(y,x);
        if(dp1[y].value + edge[i].weight > dp2[x].value)
        {
            dp2[x].value = dp1[y].value + edge[i].weight;
            dp2[x].id = y;
            if(dp1[x].value < dp2[x].value)  //dp1[i]記錄點i的最遠距離，dp2[i]記錄點i的次遠距離，
                swap(dp1[x],dp2[x]);
        }
    }
}

void DFS2(int x,int father)
{
    int i,y;
    for(i=head[x]; i!=-1; i=edge[i].next)
    {
        y = edge[i].y;
        if(y==father) continue;
        if(dp1[x].id == y) //點y是父親x的最遠距離的下一個節點
        {
            if(dp2[y].value < dp2[x].value+edge[i].weight) //，那麼看點y的次元距離能否通過父親x的其他節點更新
            {
                dp2[y].value = dp2[x].value + edge[i].weight;
                dp2[y].id = x;
                if(dp1[y].value < dp2[y].value)
                    swap(dp1[y],dp2[y]);
            }
        }
        else
        {
            if(dp2[y].value < dp1[x].value+edge[i].weight)
            {
                dp2[y].value = dp1[x].value+edge[i].weight;
                dp2[y].id = x;
                if(dp1[y].value < dp2[y].value)
                    swap(dp1[y],dp2[y]);
            }
        }
        DFS2(y,x);
    }
}

int main()
{
    int i,n;
    int x,y,k;
    while(scanf("%d",&n)!=EOF)
    {
        Init();
        for(y=2; y<=n; ++y)
        {
            scanf("%d %d",&x,&k);
            Add(x,y,k);
            Add(y,x,k);
        }
        DFS1(1,-1);
        DFS2(1,-1);
        for(i=1; i<=n; ++i)
            printf("%d\n",dp1[i].value);
    }
    return 0;
}