HDU 2196

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2196

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

参考来自http://blog.csdn.net/shuangde800/article/details/9732825的思路：

跟思路中不太相同的是，我们如下设置DP数组：

dp[i][0] : 表示以i为根的子树中的结点与i的最大距离

dp[i][1] : 表示以i为根的子树中的结点与u的次大距离（即上述思路中的secondDist）

dp[i][2] : 表示i往父亲节点方向走的最大距离

同样的，我们也做两次DFS，但是第一次的时候，我们将dp[i][0]和dp[i][1]一起算出来；

第二次DFS就可以直接用dp[i][0]和dp[i][1]。

代码如下：

#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
#define MAXN 10000+5
using namespace std;
//dp[i][0] : 表示以i为根的子树中的结点与i的最大距离
//dp[i][1] : 表示以i为根的子树中的结点与u的次大距离
//dp[i][2] : 表示i往父亲节点方向走的最大距离 
int n,dp[MAXN][3],idx[MAXN];
struct Edge{
    int u,v,w;
};
vector<Edge> child[MAXN];
void dfs1(int now)
{
    if(child[now].size()==0)
    {
        dp[now][0]=0;
        dp[now][1]=0;
        return;
    }
    for(int i=0;i<child[now].size();i++)
    {
        Edge edge=child[now][i];
        int next=edge.v;
        dfs1(next);
        if(dp[now][0] < dp[next][0]+edge.w) // ( "其某个孩子的最大"+"其与孩子的距离" ) > "最大" > "次大"
        {
            dp[now][1] = dp[now][0];
            dp[now][0] = dp[next][0] + edge.w;
            idx[now]=next;
        }
        else if(dp[now][1] < dp[next][0]+edge.w) // "最大" > ( "其某个孩子的最大"+"其与孩子的距离" ) > "次大" 
        {
            dp[now][1] = dp[next][0]+edge.w;
        }
    }
}
void dfs2(int now)
{
    for(int i=0;i<child[now].size();i++)
    {
        Edge edge=child[now][i];
        int next=edge.v;
        if(idx[now]==next) dp[next][2] = edge.w + max(dp[now][1],dp[now][2]);
        else dp[next][2] = edge.w + max(dp[now][0],dp[now][2]);
        dfs2(next);
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<=MAXN;i++) child[i].clear();
        memset(dp,0,sizeof(dp));
        memset(idx,0,sizeof(idx));
        for(int i=2;i<=n;i++)
        {
            int father,length;
            scanf("%d%d",&father,&length);
            child[father].push_back((Edge){father,i,length});
        }
        dfs1(1);
        dfs2(1);
        for(int i=1;i<=n;i++)
        {
            printf("%d
",max(dp[i][0],dp[i][2]));
        }
    }
}

总的来说，这是一道很不错的树形DP题。