Submission #525884

#	Time	Username	Problem	Language	Result	Execution time	Memory
525884		koioi.org-koosaga	Mountains and Valleys (CCO20_day1problem3)	C++17	25 / 25	6663 ms	379784 KiB

Main

#include <bits/stdc++.h>
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
using namespace std;
typedef long long lint;
typedef pair<int, int> pi;
const int MAXN = 500005;

struct edge{
	int s, e, x;
};

int n, m;
vector<int> gph[MAXN];
int par[20][MAXN], dep[MAXN];

int lca(int x, int y){
	if(dep[x] > dep[y]) swap(x, y);
	int dx = dep[y] - dep[x];
	for(int i = 0; dx; i++){
		if(dx & 1) y = par[i][y];
		dx >>= 1;
	}
	if(x == y) return x;
	for(int i = 17; i >= 0; i--){
		if(par[i][x] != par[i][y]){
			x = par[i][x];
			y = par[i][y];
		}
	}
	return par[0][x];
}

struct node{
	int lmax, rmax, opt, len;
	node():
	lmax(-1e9), rmax(-1e9), opt(-1e9), len(0) {}
	node(int x):
	lmax(x), rmax(x), opt(-1e9), len(1) {}

	node(int a, int b, int c, int e):
	lmax(a), rmax(b), opt(c), len(e) {}

	node reverse(){
		return node(rmax, lmax, opt, len);
	}
	node operator+(const node &n)const{
		return node(
			max(lmax - n.len, n.lmax),
			max(rmax, n.rmax - len),
			max({opt, n.opt, lmax + n.rmax - 1}),
			len + n.len);
	};
};

vector<int> ord;
int din[MAXN], dout[MAXN], piv;
int far[MAXN], diam[MAXN], pfar[MAXN], pdiam[MAXN];
int f[20][MAXN];
node g[20][MAXN];
vector<pi> fars[MAXN];
vector<pi> subDiams[MAXN];

int fPathMax(int x, int v){
	int ans = -1e9;
	for(int i = 0; v; i++){
		if(v & 1){
			ans = max(ans, f[i][x]);
			x = par[i][x];
		}
		v >>= 1;
	}
	return ans;
}

node gPathSum(int x, int v){
	node gs;
	for(int i = 0; v; i++){
		if(v & 1){
			gs = gs + g[i][x];
			x = par[i][x];
		}
		v >>= 1;
	}
	return gs;
}

bool in(int u, int v){
	return din[u] <= din[v] && dout[v] <= dout[u];
}

void dfs(int x, int p = -1){
	ord.push_back(x);
	din[x] = ++piv;
	for(auto &y : gph[x]){
		if(y == p) continue;
		par[0][y] = x;
		dep[y] = dep[x] + 1;
		dfs(y, x);
	}
	dout[x] = piv;
}

bool vis[MAXN];

pi dfsl(int x, int p = -1){
	pi ret(0, x);
	for(auto &y : gph[x]){
		if(y == p || vis[y]) continue;
		auto ans = dfsl(y, x);
		ans.first++;
		ret = max(ret, ans);
	}
	return ret;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> m;
	vector<edge> qry;
	for(int i = 0; i < m; i++){
		int s, e, x; cin >> s >> e >> x;
		if(x == 1){
			gph[s].push_back(e);
			gph[e].push_back(s);
		}
		else{
			qry.push_back({s, e, x});
		}
	}
	dfs(0);
	for(int i = 1; i < 20; i++){
		for(int j = 0; j < n; j++){
			par[i][j] = par[i-1][par[i-1][j]];
		}
	}
	{
		reverse(all(ord));
		for(auto &i : ord){
			for(auto &j : gph[i]){
				if(j == par[0][i]) continue;
				far[i] = max(far[i], far[j] + 1);
				diam[i] = max({diam[i], diam[j], far[j] + 1});
				fars[i].emplace_back(far[j], j);
			}
			sort(all(fars[i]));
			reverse(all(fars[i]));
			if(sz(fars[i]) >= 2) diam[i] = max(diam[i], (int)fars[i][0].first + (int)fars[i][1].first + 2);
		}
		reverse(all(ord));
		for(auto &i : ord){
			if(i == 0) continue;
			for(auto &j : fars[par[0][i]]){
				if(j.second == i) continue;
				pfar[i] = j.first + 1;
				break;
			}
			fars[i].emplace_back(pfar[i], par[0][i]);
			sort(all(fars[i]));
			reverse(all(fars[i]));
		}	
		for(auto &i : ord){
			if(i == 0) continue;
			for(auto &[d, v] : subDiams[par[0][i]]){
				if(v == i) continue;
				pdiam[i] = max(pdiam[i], (int)d);
				break;
			}
			{
				int new_diam = 0;
				int cnt = 0;
				for(auto &[d, v] : fars[par[0][i]]){
					if(v == i) continue;
					new_diam += d + 1;
					cnt += 1;
					if(cnt == 2) break;
				}
				pdiam[i] = max(pdiam[i], new_diam);
			}
			for(auto &j : gph[i]){
				if(j == par[0][i]) subDiams[i].emplace_back(pdiam[i], j);
				else subDiams[i].emplace_back(diam[j], j);
			}
			sort(all(subDiams[i]));
			reverse(all(subDiams[i]));
		}
		for(int i = 0; i < n; i++){
			if(dep[i] >= 2){
				int cnt = 0;
				int gi = 0;
				for(auto &[d, v] : fars[par[0][i]]){
					if(v == i || v == par[1][i]) continue;
					gi = max(gi, (int)d + 1);
					f[0][i] += d + 1;
					cnt++;
					if(cnt == 2) break;
				}
				for(auto &[d, v] : subDiams[par[0][i]]){
					if(v == i || v == par[1][i]) continue;
					f[0][i] = max(f[0][i], (int)d);
					break;
				}
				g[0][i] = node(gi);
			}
		}
		for(int i = 1; i < 20; i++){
			for(int j = 0; j < n; j++){
				f[i][j] = max(f[i-1][j], f[i-1][par[i-1][j]]);
				g[i][j] = g[i-1][j] + g[i-1][par[i-1][j]];
			}
		}
	}
	int ans = 2 * n - 2 - diam[0];
	for(auto &x : qry){
		int l = lca(x.s, x.e);
		int plen = dep[x.s] + dep[x.e] - 2 * dep[l] + 1;
		int maxPath = -1e9;
		if(dep[x.s] >= dep[l] + 2){
			maxPath = max(maxPath, fPathMax(x.s, dep[x.s] - dep[l] - 1) - 3 + plen);
		}
		if(dep[x.e] >= dep[l] + 2){
			maxPath = max(maxPath, fPathMax(x.e, dep[x.e] - dep[l] - 1) - 3 + plen);
		}
		if(l != x.s){
			maxPath = max(maxPath, diam[x.s] - 3 + plen);
		}
		if(l != x.e){
			maxPath = max(maxPath, diam[x.e] - 3 + plen);
		}
		{
			int new_diam = 0;
			int cnt = 0;
			// passes through l
			for(auto &[d, v] : fars[l]){
				if((in(v, x.s) || in(v, x.e)) && v != par[0][l]) continue;
				new_diam += d + 1;
				cnt += 1;
				if(cnt == 2) break;
			}
			for(auto &[d, v] : subDiams[l]){
				if((in(v, x.s) || in(v, x.e)) && v != par[0][l]) continue;
				new_diam = max(new_diam, (int)d);
				break;
			}
			maxPath = max(maxPath, new_diam - 3 + plen);
		}
		// TODO: qn.
		node valup, valdown;
		if(x.s != l) valup = valup + node(far[x.s]);
		if(x.e != l) valdown = valdown + node(far[x.e]);
		if(dep[x.s] >= dep[l] + 2){
			valup = valup + gPathSum(x.s, dep[x.s] - dep[l] - 1);
		}
		if(dep[x.e] >= dep[l] + 2){
			valdown = valdown + gPathSum(x.e, dep[x.e] - dep[l] - 1);
		}
		{
			int dist = 0;
			for(auto &[d, v] : fars[l]){
				if((in(v, x.s) || in(v, x.e)) && v != par[0][l]) continue;
				dist = max(dist, (int)d + 1);
				break;
			}
			valup = valup + node(dist);
		}
		node val = valup + valdown.reverse();
		maxPath = max(maxPath, val.opt + val.len - 1);
		ans = min(ans, 2 * n - 4 + x.x - maxPath);
	}
	cout << ans << endl;
}

• Subtask 1: 1 / 1
• Subtask 2: 2 / 2
• Subtask 3: 2 / 2
• Subtask 4: 6 / 6
• Subtask 5: 2 / 2
• Subtask 6: 3 / 3
• Subtask 7: 5 / 5
• Subtask 8: 4 / 4