hdu 2196 Computer（树形dp)

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

题意：在一个每条边都有权值的树中，求每个顶点能走的最大距离;

思路：两次dfs,以1为根结点，一次从叶子到根结点，求出最远距离和次远距离。另一次从根进行dfs.

(res[]为经过父节点能走最长距离）

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#define sum 10010
using namespace std;
int f[sum] ={0};
int g[sum] ={0};
int biao[sum]= {0};
int res[sum] = {0};
int n = 0;
int a = 0;
int b = 0;
vector <int> son[sum],num[sum];
void init()
{
    memset(f , 0, sizeof(f));
    memset(g , 0, sizeof(g));
    memset(biao , -1, sizeof(biao));
    memset(res , 0, sizeof(res));
    for(int i = 1;i <= n; ++i)
    {
    son[i].clear();
    num[i].clear();
    }
}
int dfs(int nn)
{
    int nu = son[nn].size();
    if( 0 == nu)
    return 0;
    for(int i = 0; i < nu ;++i)
    {
        int temp = dfs(son[nn][i]);
        if(temp + num[nn][i] > g[nn])
        {
        if(temp + num[nn][i] >= f[nn])
        {
            g[nn] = f[nn];
        f[nn] = temp + num[nn][i];
        biao[nn] = i;
        }
        else
        {
         g[nn] = temp + num[nn][i];
        }
        }
    }
    return f[nn];
}
int max1(int a,int b)
{
    if(a > b)
    return a;
    return b;
}
void dfs1(int nn)
{
    int nu = son[nn].size();
    for(int i = 0; i < nu ;++i)
    {

       if(i == biao[nn])
       {
           res[son[nn][i]] =max1(res[nn],g[nn]) + num[nn][i]  ;
       }

       else
       res[son[nn][i]] = max1(res[nn],f[nn]) + num[nn][i] ;
       dfs1(son[nn][i]);
    }
}
int main()
{
    while(scanf("%d",&n) != EOF)
    {
        init();
        for(int i = 2;i <= n; ++i)
        {
           scanf("%d%d",&a,&b);
           son[a].push_back(i);
           num[a].push_back(b);
        }
        f[1] = dfs(1);
        dfs1(1);
        res[1] = f[1];
        for(int i = 1; i <= n; ++i)
        printf("%d\n",max1(f[i],res[i]));
    }
    return 0;
}