Computer （HDU - 2196，树形 DP - 二次扫描与换根法）

一.题目链接：

HDU-2196

二.题目大意：

给一颗无根树，求每个节点所能到达节点的最大距离.

三.分析：

感觉换根好难搞啊，想不清该维护哪些量，额我好笨啊...

附一个大佬讲解，看完就懂了~

还是要多做一些换根dp呀 (ง •̀_•́)ง

四.代码实现：

#include <bits/stdc++.h>
using namespace std;

const int M = (int)1e4;
const int inf = 0x3f3f3f3f;

int cnt;
int head[M + 5];
struct node
{
    int v, w, nx;
} Edge[M * 2 + 5];

int mx[M + 5], mxi[M + 5];
int sm[M + 5];
int dp[M + 5][2];

void init(int n)
{
    cnt = 0;
    for(int i = 1; i <= n; ++i)
    {
        head[i] = -1;
        mx[i] = sm[i] = -inf;
        dp[i][0] = dp[i][1] = 0;
    }
}

void add(int u, int v, int w)
{
    Edge[cnt].v = v;
    Edge[cnt].w = w;
    Edge[cnt].nx = head[u];
    head[u] = cnt++;
}

void dfs1(int u, int fa)
{
    for(int i = head[u]; ~i; i = Edge[i].nx)
    {
        int v = Edge[i].v;
        if(v == fa)
            continue;
        dfs1(v, u);
        if(mx[u] <= mx[v] + Edge[i].w)
        {
            sm[u] = mx[u];
            mx[u] = mx[v] + Edge[i].w;
            mxi[u] = v;
        }
        else if(sm[u] < mx[v] + Edge[i].w)
        {
            sm[u] = mx[v] + Edge[i].w;
        }
    }
    if(mx[u] == -inf)
        mx[u] = 0;
}

void dfs2(int u, int fa)
{
    for(int i = head[u]; ~i; i = Edge[i].nx)
    {
        int v = Edge[i].v;
        if(v == fa) continue;
        dp[v][0] = mx[v];
        if(mxi[u] == v)
            dp[v][1] = max(sm[u], dp[u][1]) + Edge[i].w;
        else
            dp[v][1] = max(mx[u], dp[u][1]) + Edge[i].w;
        dfs2(v, u);
    }
}

int main()
{
//    freopen("input.txt", "r", stdin);
    int n;
    while(~scanf("%d", &n))
    {
        init(n);
        for(int u = 2, v, w; u <= n; ++u)
        {
            scanf("%d %d", &v, &w);
            add(u, v, w);
            add(v, u, w);
        }
        dfs1(1, 0);
        dp[1][0] = mx[1];
        dfs2(1, 0);
        for(int i = 1; i <= n; ++i)
            printf("%d\n", max(dp[i][0], dp[i][1]));
    }
    return 0;
}