HDU-2196-Computer(树上DP)

链接：

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

题意：

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.

思路：

可以一眼看出可以用换根.
但是无法处理是否上一个节点的最长距离正好选了某个子节点，看了题解发现可以去记录最长的和次长的，长见识了。。
一次DFS求出有向的每个点往下的最长距离和次长距离，然后再一次DFS，求出往父节点扩展的长度。

代码：

// #include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 1e9;
const int MAXN = 1e4+10;

struct Node
{
    int x;
    int v;
};
vector<Node> G[MAXN];
int Dp1[MAXN], Dp2[MAXN], Dp3[MAXN];
int Max[MAXN];
int n;

void Dfs1(int x, int pre)
{
    Dp1[x] = Dp2[x] = 0;
    for (int i = 0;i < G[x].size();i++)
    {
        int node = G[x][i].x;
        if (node == pre)
            continue;
        Dfs1(node, x);
        if (Dp1[node]+G[x][i].v > Dp1[x])
        {
            Dp1[x] = Dp1[node]+G[x][i].v;
            Max[x] = node;
        }
    }
    for (int i = 0;i < G[x].size();i++)
    {
        int node = G[x][i].x;
        if (node == Max[x])
            continue;
        Dp2[x] = max(Dp2[x], Dp1[node]+G[x][i].v);
    }
}

void Dfs2(int x, int pre)
{
    for (int i = 0;i < G[x].size();i++)
    {
        int node = G[x][i].x;
        if (node == pre)
            continue;
        if (node != Max[x])
            Dp3[node] = max(Dp1[x]+G[x][i].v, Dp3[x]+G[x][i].v);
        else
            Dp3[node] = max(Dp2[x]+G[x][i].v, Dp3[x]+G[x][i].v);
        Dfs2(node, x);
    }
}

int main()
{
    // freopen("test.in", "r", stdin);
    int t, cas = 0;
    // scanf("%d", &t);
    while(~scanf("%d", &n))
    {
        for (int i = 1;i <= n;i++)
            G[i].clear();
        memset(Dp1, 0, sizeof(Dp1));
        memset(Dp2, 0, sizeof(Dp2));
        int a, l;
        for (int i = 2;i <= n;i++)
        {
            scanf("%d%d", &a, &l);
            G[a].push_back(Node{i, l});
            G[i].push_back(Node{a, l});
        }
        Dfs1(1, -1);
        Dfs2(1, -1);
        for (int i = 1;i <= n;i++)
            printf("%d
", max(Dp1[i], Dp3[i]));
    }

    return 0;
}