HDU 2196 Compute --树形dp

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35047    Accepted Submission(s): 5633

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

题意：给你一棵树，求每个节点的最远对点的距离。

题目看似有点难度，但我们想一想两遍dfs求树的直径的过程，第一遍dfs找的是随意一个点的最远对点，它一定是树的直径的端点，那么我们就可以猜想一下是不是每个点的最远对点都是树的直径的端点呢？答案是肯定的。那么我们就可以先把树的直径找出来然后分别求两个端点到所有的点的距离，取最大值即可。

代码如下：

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#define maxn 10005
using namespace std;

struct edge
{
    int next;
    int to;
    int dis;
}g[maxn<<1];

inline int read()
{
    char c=getchar();
    int res=0,x=1;
    while(c<'0'||c>'9')
    {
        if(c=='-')
        x=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9')
    {
        res=res*10+(c-'0');
        c=getchar();
    }
    return x*res;
}

int n,aa,bb,num,root,ans;
int last[maxn],d[maxn],dp[maxn],dp1[maxn];

inline void add(int from,int to,int dis)
{
    g[++num].next=last[from];
    g[num].to=to;
    g[num].dis=dis;
    last[from]=num;
}

void dfs(int x)
{
    d[x]=1;
    for(int i=last[x];i;i=g[i].next)
    {
        int v=g[i].to;
        if(!d[v])
        {
            dp[v]=dp[x]+g[i].dis;
            dfs(v);
        }
    }
}

void dfs1(int x)
{
    d[x]=1;
    for(int i=last[x];i;i=g[i].next)
    {
        int v=g[i].to;
        if(!d[v])
        {
            dp[v]=dp[x]+g[i].dis;
            dfs1(v);
        }
    }
}

void dfs2(int x)
{
    d[x]=1;
    for(int i=last[x];i;i=g[i].next)
    {
        int v=g[i].to;
        if(!d[v])
        {
            dp1[v]=dp1[x]+g[i].dis;
            dfs2(v);
        }
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;num=0;
        memset(last,0,sizeof(last));
        memset(dp,0,sizeof(dp));
        memset(d,0,sizeof(d));
        memset(dp1,0,sizeof(dp1));
        for(int i=2;i<=n;i++)
        {
            aa=read();bb=read();
            add(i,aa,bb);
            add(aa,i,bb);
        }
        dfs(1);
        for(int i=1;i<=n;i++)
        {
            if(dp[i]>ans)
            {
                ans=dp[i];
                root=i;
            }
        }
        memset(dp,0,sizeof(dp));
        memset(d,0,sizeof(d));
        dfs1(root);
        ans=0;
        memset(d,0,sizeof(d));
        for(int i=1;i<=n;i++)
        {
            if(dp[i]>ans)
            {
                ans=dp[i];
                root=i;
            }
        }
        dfs2(root);
        for(int i=1;i<=n;i++)
        {
            printf("%d
",max(dp[i],dp1[i]));
        }
    }
    return 0;
}

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                                                                                                                              2019-01-18  07:49:19