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：

参考博客：逍遥丶綦

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

细节还是挺多的。而且借助队列和前后缀处理太强啦。

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;
}