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);

}

巴特西

Codeforces 633F 树的直径/树形DP

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