換根dp

換根dp

描述：

題目初始不給出根節點，需要以每一個節點為根節點遍歷，打到某一個根節點的最佳情況。

暴力的思路是迴圈每個節點做為根節點，同時在遍歷整個樹，時間複雜度是O(n^2)

換根dp主要是將時間複雜度降到O(n)，在根節點切換時，直接透過一些已經計算過的資料在O(1)就能得到另一個根的結果。

一般是透過二次掃描，第一次dfs獲得預處理資料，第二次dfs進行根節點的切換。

例題

E - Minimize Sum of Distances

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

#define int long long
const int maxn = 1e5 + 5;
vector<int> adj[maxn];

void solve() {
	int n;
	cin >> n;
	
	for (int i = 0; i < n - 1; i++) {
		int a, b;
		cin >> a >> b;
		adj[a].push_back(b);
		adj[b].push_back(a);
	}
	vector<int> v(n + 1);
	for (int i = 1; i <= n; i++) {
		cin >> v[i];
	}
			
	vector<int> fa(n + 1);
	vector<int> p (n + 1);
	vector<int> cnt(n + 1);
	function<void(int)> dfs =[&] (int x) {
		cnt[x] = v[x];
		p[x] = 0;
		
		for (auto it : adj[x]) {
			if(it != fa[x]) {
				fa[it] = x;
				dfs(it);
				cnt[x] += cnt[it];
				p[x] += cnt[it] + p[it];
			}
		}	
	};
	dfs(1);
	
	int res = p[1];	
	vector<int> f(n + 1);
	f[1] = p[1];
	function<void(int, int)> dfs2=[&](int u, int v) {
		if(v != 1) {
			f[v] = f[u] + cnt[1] - cnt[v] - cnt[v];
			res = min(res, f[v]);
		}
		for (auto it : adj[v]) {
			if(it != fa[v]) {
				dfs2(v, it);
			}
		}
	};
	
	dfs2(1, 1);
	cout << res << endl;
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int T = 1;

	while(T--) {
		solve();
	}

	return 0;
}

積蓄程度

卡map

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

#define int long long
const int maxn = 2e5 + 5;
vector<pair<int, int>> adj[maxn];

void solve() {
	int n;
	cin >> n;
	
	for (int i = 1; i <= n; i++) {
		adj[i].clear();
	}
	map<pair<int, int>, int> mp;
	for (int i = 0; i < n - 1; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		
		adj[a].push_back({b, c});
		adj[b].push_back({a, c});
	}
		
	vector<int> p(n + 1);
	vector<int> f(n + 1);
		
	function<int(int, int, int)> dfs1 =[&] (int u, int fa, int e) {
		p[u] = 0;
		if(adj[u].size() == 1 && adj[u][0].first == fa) {
			return e;
		}
		
		for (auto it : adj[u]) {
			if(it.first != fa) {
				p[u] += dfs1(it.first, u, it.second);
			}
		}
		
		return min(p[u], e);
	};
	
	p[1] = dfs1(1, 0, 1e18);
	f[1] = p[1];
	
	int ans = p[1];
	function<void(int, int, int e)> dfs2 =[&] (int u, int v, int e) {
		if(v != 1) {
			if(adj[u].size() == 1) {
				f[v] = p[v] + e;
			}
			else {
				f[v] = min(f[u] - min(p[v], e), e) + p[v];
			}
			ans = max(f[v], ans);
		}
		for (auto it : adj[v]) {
			if(it.first != u) {
				dfs2(v, it.first, it.second);
			}
		}
	};
	
	dfs2(0, 1, 0);
	
	cout << ans << endl;
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int T = 1;
	cin >> T;
	
	while(T--) {
		solve();
	}

	return 0;
}

P3478 [POI2008] STA-Station

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

#define int long long
const int maxn = 1e6 + 5;
vector<int> adj[maxn];

void solve() {
	int n;
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int a, b;
		cin >> a >> b;
		adj[a].push_back(b);
		adj[b].push_back(a);
	}	
		
	vector<int> fa(n + 1);
	vector<int> p(n + 1);
	vector<int> f(n + 1);
	vector<int> cnt(n + 1);
	
	function<void(int)> dfs1 =[&](int u) {
		p[u] = 1;
		cnt[u] = 1;
		if(adj[u].size() == 1 && adj[u][0] == fa[u]) {
			return ;
		}
		
		for (auto it : adj[u]) {
			if(it != fa[u]) {
				fa[it] = u;
				dfs1(it);
				p[u] += p[it] + cnt[it];
				cnt[u] += cnt[it];
			}
		}	
	};
	
	dfs1(1);
	int res = p[1];
	int id = 1;
	f[1] = p[1];
	function<void(int, int v)> dfs2=[&](int u, int v) {
		if(v != 1) {
			f[v] = f[u] - cnt[v] + cnt[1] - cnt[v];
			if(f[v] > res) {
				id = v;
				res = f[v];
			}
		}
		
		for (auto it : adj[v]) {
			if(it != fa[v]) {
				dfs2(v, it);
			}
		}
	};
	
	dfs2(1, 1);
	cout << id << endl;
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int T = 1;
	
	while(T--) {
		solve();
	}

	return 0;
}

P3047 [USACO12FEB] Nearby Cows G

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

#define int long long
	
const int maxn = 1e5 + 5;
vector<int> adj[maxn];
int dp[maxn][25];

void solve() {
	int n, k;
	cin >> n >> k;
	
	vector<int> v(n + 1);
	for (int i = 0; i < n - 1; i++) {
		int a, b;
		cin >> a >> b;
		adj[a].push_back(b);
		adj[b].push_back(a);
	}
	
	for (int i = 1; i <= n; i++) {
		cin >> v[i];
		for (int j = 0; j <= k; j++) {
			dp[i][j] = 0;
		}
	}
	
	function<void(int, int)> dfs1 =[&] (int u, int fa) {
			dp[u][0] = v[u];
			for (auto it : adj[u]) {
				if(it != fa) {
					dfs1(it, u);
					for (int i = 1; i <= k; i++) {
						dp[u][i] += dp[it][i - 1];
					}
				}
			}
	};
	
	dfs1(1, 0);
	vector<int> ans(n + 1);
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < k; j++) {
			dp[i][j + 1] += dp[i][j];
		}
	}
	
	vector<int> vt;
	function<void(int, int)> dfs2 =[&] (int u, int fa) {
		int res = 0;
		vt.push_back(u);
		for (int i = vt.size() - 2, j = 0; i >= 0 && j < k; i--, j++) {
			res += dp[vt[i]][k - j - 1] - dp[vt[i + 1]][k - j - 2];
		}
		ans[u] = res + dp[u][k];
		for (auto it : adj[u]) {
			if(it != fa) {
				dfs2(it, u);
			}
		}
		vt.pop_back();	
	};
	
	dfs2(1, 0);
	for (int i = 1; i <= n; i++) {
		cout << ans[i] << endl;
	}
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int T = 1;
//	cin >> T;
	
	while(T--) {
		solve();		
	}
	
	return 0;
}