Codeforces 633F 树的直径/树形DP

题意：有两个小孩玩游戏，每个小孩可以选择一个起始点，并且下一个选择的点必须和自己选择的上一个点相邻，问两个选的点权和的最大值是多少？

思路：首先这个问题可以转化为求树上两不相交路径的点权和的最大值，对于这种问题，我们有两种想法：

1：树的直径，受之前HDU多校的那道题的启发，我们先找出树的直径，然后枚举保留直径的哪些部分，去找保留这一部分的最优解，去更新答案。

代码：

#include <bits/stdc++.h>
#define INF 1e18
#define LL long long
using namespace std;
const int maxn = 100010;
vector<int> G[maxn];
LL tot;
int now, f[maxn];
LL d[maxn], val[maxn], sum[maxn];
bool v[maxn], v1[maxn];
LL mx[maxn];
void add(int x, int y) {
	G[x].push_back(y);
	G[y].push_back(x);
}
void dfs(int x, int fa, LL sum) {
	sum += val[x];
	v1[x] = 1;
	f[x] = fa;
	if(sum > tot) {
		now = x;
		tot = sum;
	}
	for (auto y : G[x]) {
		if(y == fa || v[y]) continue;
		dfs(y, x, sum);
	}
}
void dfs1(int x, int fa) {
	d[x] = val[x];
	LL tmp = 0;
	for (auto y : G[x]) {
		if(y == fa || v[y]) continue;
		dfs1(y, x);
		tmp = max(tmp, d[y]);
	}
	d[x] += tmp;
	return;
}
vector<int> a;
int main() {
	int n, x, y;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &val[i]);
	}
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &x, &y);
		add(x, y);
	}
	LL ans = 0;
	tot = 0, now = 0;
	dfs(1, 0, 0);
	tot = 0;
	dfs(now, 0, 0);
	for (int i = now; i; i = f[i]) {
		a.push_back(i);
		sum[a.size()] = sum[a.size() - 1] + val[i];
		v[i] = 1;
	}
	for (auto y : a) {
		dfs1(y, 0);
	}
	memset(v1, 0, sizeof(v1));
	for (int i = 1; i <= n; i++) {
		if(v[i] == 1 || v1[i] == 1) continue;
		tot = now = 0;
		dfs(i, 0, 0);
		tot = 0;
		dfs(now, 0, 0);
		ans = max(ans, tot + sum[a.size()]);
	}
	mx[a[a.size() - 1]] = d[a[a.size() - 1]];
	for (int i = a.size() - 2; i >= 1; i--) {
		mx[a[i]] = max(mx[a[i + 1]], sum[a.size()] - sum[i] + d[a[i]] - val[a[i]]);
	}
	for (int i = 0; i < a.size() - 1; i++) {
		ans = max(ans, d[a[i]] + sum[i] + mx[a[i + 1]]);
	}
	printf("%lld\n", ans);
}

思路2：树形DP，我们对于每个点保留从它到叶子节点的最长路径和次长路径，以及以它为根的子树中的最长路径。dp完之后，我们对于每个节点，枚举选哪棵子树中的节点作为一条路径，然后去寻找另一条最长路径。可以用前缀最大值和后缀最大值去优化，以及需要注意需要考虑父节点方向对答案的影响。

代码：

#include <bits/stdc++.h>
#define LL long long
#define pll pair<LL, LL>
using namespace std;
const int maxn = 100010;
vector<int> G[maxn];
LL mx_path[maxn], mx_dis[maxn];
LL lmx_path[maxn], rmx_path[maxn];
pll lmx_dis[maxn], rmx_dis[maxn];
LL val[maxn];
LL ans = 0;
void add(int x, int y) {
	G[x].push_back(y);
	G[y].push_back(x);
}
void dfs(int x, int fa) {
	vector<LL> tmp(3);
	for (auto y : G[x]) {
		if(y == fa) continue;
		dfs(y, x);
		tmp[0] = mx_dis[y];
		sort(tmp.begin(), tmp.end());
		mx_path[x] = max(mx_path[y], mx_path[x]);
	}
	mx_dis[x] = tmp[2] + val[x];
	mx_path[x] = max(mx_path[x], tmp[1] + tmp[2] + val[x]);
	return;
}
int tot, a[maxn];
struct node {
	int x, fa;
	LL mx_p, mx_d;
};
queue<node> q;
void solve() {
	q.push((node){1, 0, 0, 0});
	while(q.size()) {
		node tmp = q.front();
		q.pop();
		tot = 0;
		int x = tmp.x, fa = tmp.fa;
		tot = 0;
		for (int i = 0; i < G[x].size(); i++) {
			if(G[x][i] == fa) continue;
			a[++tot] = G[x][i];
		}
		rmx_path[tot + 1] = rmx_path[tot + 2] = 0;
		rmx_dis[tot + 1] = rmx_dis[tot + 2] = make_pair(0, 0);
		for (int i = 1; i <= tot; i++) {
			lmx_path[i] = max(lmx_path[i - 1], mx_path[a[i]]);
			lmx_dis[i] = lmx_dis[i - 1];
			if(mx_dis[a[i]] >= lmx_dis[i].first) {
				lmx_dis[i].second = lmx_dis[i].first;
				lmx_dis[i].first = mx_dis[a[i]];
			} else if(mx_dis[a[i]] >  lmx_dis[i].second) {
				lmx_dis[i].second = mx_dis[a[i]];
			}
		}
		for (int i = tot; i >= 1; i--) {
			rmx_path[i] = max(rmx_path[i + 1], mx_path[a[i]]);
			rmx_dis[i] = rmx_dis[i + 1];
			if(mx_dis[a[i]] >= rmx_dis[i].first) {
				rmx_dis[i].second = rmx_dis[i].first;
				rmx_dis[i].first = mx_dis[a[i]];
			} else if(mx_dis[a[i]] >  rmx_dis[i].second) {
				rmx_dis[i].second = mx_dis[a[i]];
			}
		}
		LL tmp1 = 0;
//		printf("%d\n", x);
//		for (int i = 1; i <= tot; i++) {
//			printf("%lld %lld\n", lmx_path[i], rmx_path[i]);
//		}
		for (int i = 1; i <= tot; i++) {
			LL tmp2 = 0;
			tmp2 = max(tmp2, lmx_path[i - 1]);
			tmp2 = max(tmp2, rmx_path[i + 1]);
			tmp2 = max(tmp2, tmp.mx_p);
			tmp2 = max(tmp2, val[x] + lmx_dis[i - 1].first + tmp.mx_d);
			tmp2 = max(tmp2, val[x] + rmx_dis[i + 1].first + tmp.mx_d);
			tmp2 = max(tmp2, val[x] + lmx_dis[i - 1].first + lmx_dis[i - 1].second);
			tmp2 = max(tmp2, val[x] + rmx_dis[i + 1].first + rmx_dis[i + 1].second);
			tmp2 = max(tmp2, val[x] + lmx_dis[i - 1].first + rmx_dis[i + 1].first);
			tmp2 = max(tmp2, val[x] + tmp.mx_d + max(lmx_dis[i - 1].first, rmx_dis[i + 1].first));
//			printf("%lld ", tmp.mx_d);
//				printf("%lld ", tmp2);
			q.push((node){a[i], x, tmp2, val[x] + max(max(lmx_dis[i - 1].first, rmx_dis[i + 1].first), tmp.mx_d)});
			tmp2 += mx_path[a[i]];
//			printf("%lld\n", tmp2);
			tmp1 = max(tmp1, tmp2);
		}
//		printf("\n");
		ans = max(ans, tmp1);
		ans = max(ans, tmp.mx_p + mx_path[x]);
	}
}
int main() {
	int n, x, y;
//	freopen("633Fin.txt", "r" , stdin);
//	freopen("out1.txt", "w", stdout);
	scanf("%d" , &n);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &val[i]);
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &x, &y);
		add(x, y);
	}
	dfs(1, 0);
	solve();
	printf("%lld\n", ans);
}