Codeforces 633 F The Chocolate Spree(树形dp，两条不相交链节点权值和最大)

两条不相交链节点权值和最大)"/>

Codeforces 633 F The Chocolate Spree(树形dp，两条不相交链节点权值和最大)

题目链接：
Codeforces 633 F The Chocolate Spree
题意：
给一个n个节点的树和n−1条边，每个点有一个权值，从树中选择两条不相交的链(无公共节点)使得两条链上节点权值和最大？
数据范围：n≤105,value[i]≤109
分析：

先dfs处理出每个节点向其子树最多可以得到的链的最大权值和down[v]（必须由v节点向下）和在子树中最长链的权值best[v]（可以由一棵子树从下往上经过v节点然后转向另一棵子树，也可以是某棵子树的内部的最长链）。
然后借助队列遍历每个u和其儿子v“截断”这条边（以这条边为分界线），求的两棵树的内部最长链。这时就相当于把v节点和u以及u的其余儿子节点都分开了。我们先求u及u除了v以外的儿子节点所形成的树的最长链，记为outside。
并记up[u]为u向上可获得的最长链（不包括u节点），predown[]和
ppredown[]为u的儿子的down[]数组的前缀最大和次大，prebest[]为儿子的前缀最大，sufdown[],ssufdown[],sufbest[]分别为相应的后缀down[]数组最大和次大，best[]数组最大。
考虑outside可以获得的组合，当前枚举到第i个儿子：

up[u]+value[u]+max(predown[i−1],sufdown[i+1])
value[u]+predown[i−1]+sufdown[i+1]
value[u]+predown[i−1]+ppredown[i−1]
value[u]+sufdown[i+1]+ssufdown[i+1]
max(prebest[i−1],sufbest[i+1])
更新：

ans=max(ans,outside+best[v]),第i个儿子是v
up[v]=value[u]+max(up[u],max(predown[i−1],sufdown[i+1]));up[v]不包括v节点权值
借助队列和前后缀处理
时间复杂度：O(n)

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <climits>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
const int MAX_N = 100010;int n, total;
int head[MAX_N];
ll value[MAX_N], best[MAX_N], down[MAX_N];
ll prebest[MAX_N], sufbest[MAX_N], predown[MAX_N], ppredown[MAX_N];
ll sufdown[MAX_N], ssufdown[MAX_N], up[MAX_N];struct Edge {int to, next;
}edge[MAX_N * 2];inline void AddEdge(int from, int to)
{edge[total].to = to, edge[total].next = head[from];head[from] = total++;
}void dfs(int u, int p)
{ll Max = 0, MMax = 0;for (int i = head[u]; i != -1; i = edge[i].next) {int v = edge[i].to;if (v == p) continue;dfs(v, u);if (down[v] > Max) {MMax = Max;Max = down[v];} else if (down[v] > MMax) {MMax = down[v];}best[u] = max(best[u], best[v]); }   down[u] = Max + value[u];best[u] = max(best[u], Max + MMax + value[u]); // best可以经过根节点
}void solve()
{ll ans = 0;queue<pair<int, int> > que;que.push(make_pair(1, 0));while (!que.empty()) {pair<int, int> cur = que.front();int u = cur.first, p = cur.second;que.pop();vector<int> child;child.push_back(0);for (int i = head[u]; i != -1; i = edge[i].next) {int v = edge[i].to;if (v != p) child.push_back(v);}int size = child.size();// 记录down数组前缀最大和次大，best数组前缀最大prebest[0] = predown[0] = ppredown[0] = 0;for (int i = 1; i < size; ++i) {int v = child[i];prebest[i] = max(prebest[i - 1], best[v]);predown[i] = predown[i - 1], ppredown[i] = ppredown[i - 1];if (down[v] > predown[i]) {ppredown[i] = predown[i];predown[i] = down[v];} else if (down[v] > ppredown[i]) {ppredown[i] = down[v];}}// 记录down数组后缀最大和次大，best数组后缀最大sufbest[size] = sufdown[size] = ssufdown[size] = 0;for (int i = size - 1; i >= 1; --i) {int v = child[i];sufbest[i] = max(sufbest[i + 1], best[v]);sufdown[i] = sufdown[i + 1], ssufdown[i] = ssufdown[i + 1];if (down[v] > sufdown[i]) {ssufdown[i] = sufdown[i];sufdown[i] = down[v];} else if (down[v] > ssufdown[i]) {ssufdown[i] = down[v];}}// 遍历每个儿子，并且每个儿子的best是一条链for (int i = 1; i < size; ++i) {int v = child[i];ll outside = up[u] + value[u] + max(predown[i - 1], sufdown[i + 1]);outside = max(outside, value[u] + predown[i - 1] + ppredown[i - 1]);outside = max(outside, value[u] + sufdown[i + 1] + ssufdown[i + 1]);outside = max(outside, value[u] + predown[i - 1] + sufdown[i + 1]);outside = max(outside, max(prebest[i - 1], sufbest[i + 1]));ans = max(ans, outside + best[v]);}// 更新儿子的up并将儿子压进队列for (int i = 1; i < size; ++i) {int v = child[i];up[v] = value[u] + max(up[u], max(predown[i - 1], sufdown[i + 1]));que.push(make_pair(v, u));}       }printf("%lld\n", ans);
}int main()
{while (~scanf("%d", &n)) {for (int i = 1; i <= n; ++i) { scanf("%lld", &value[i]); }memset(head, -1, sizeof(head));total = 0;for (int i = 1; i < n; ++i) {int u, v;scanf("%d%d", &u, &v);AddEdge(u, v);AddEdge(v, u);}memset(best, 0, sizeof(best));memset(down, 0, sizeof(down));memset(up, 0, sizeof(up));dfs(1, 0);solve();}return 0;
}