#386578 (C++14) No.898 tri-βutree

提出ソース
結果

問題	No.898 tri-βutree
ユーザー	satanic
提出日時	2019-10-04 21:33:50
言語	C++14 (gcc 13.3.0 + boost 1.87.0)
結果	AC
実行時間	387 ms / 4,000 ms
コード長	7,289 bytes
コンパイル時間	1,746 ms
コンパイル使用メモリ	138,716 KB
実行使用メモリ	36,672 KB
最終ジャッジ日時	2024-11-08 21:53:55
合計ジャッジ時間	9,527 ms
ジャッジサーバーID （参考情報）	judge3 / judge1
このコードへのチャレンジ
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ファイルパターン	結果
sample	AC * 1
other	AC * 21
権限があれば一括ダウンロードができます
ソースコード

raw source code
// need
#include <iostream>
#include <algorithm>
// data structure
#include <bitset>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <complex>
//#include <deque>
#include <valarray>
#include <unordered_map>
#include <unordered_set>
#include <array>
// etc
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>

// input
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
#define VAR(type, ...)type __VA_ARGS__;MACRO_VAR_Scan(__VA_ARGS__);
template<typename T> void MACRO_VAR_Scan(T& t) { std::cin >> t; }
template<typename First, typename...Rest>void MACRO_VAR_Scan(First& first, Rest& ...rest) { std::cin >> first; MACRO_VAR_Scan(rest...); }
#define VEC_ROW(type, n, ...)std::vector<type> __VA_ARGS__;MACRO_VEC_ROW_Init(n, __VA_ARGS__); for(int w_=0; w_<n; ++w_){MACRO_VEC_ROW_Scan(w_, __VA_ARGS__);}
template<typename T> void MACRO_VEC_ROW_Init(int n, T& t) { t.resize(n); }
template<typename First, typename...Rest>void MACRO_VEC_ROW_Init(int n, First& first, Rest& ...rest) { first.resize(n); MACRO_VEC_ROW_Init(n, rest...); }
template<typename T> void MACRO_VEC_ROW_Scan(int p, T& t) { std::cin >> t[p]; }
template<typename First, typename...Rest>void MACRO_VEC_ROW_Scan(int p, First& first, Rest& ...rest) { std::cin >> first[p]; MACRO_VEC_ROW_Scan(p, rest...); }
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define MAT(type, c, m, n) std::vector<std::vector<type>> c(m, std::vector<type>(n));for(auto& R:c)for(auto& w:R)std::cin>>w;
// output
#define OUT(dist) std::cout<<(dist);
#define FOUT(n, dist) std::cout<<std::fixed<<std::setprecision(n)<<(dist);
#define SOUT(n, c, dist) std::cout<<std::setw(n)<<std::setfill(c)<<(dist);
#define SP std::cout<<" ";
#define TAB std::cout<<"\t";
#define BR std::cout<<"\n";
#define SPBR(w, n) std::cout<<(w + 1 == n ? '\n' : ' ');
#define ENDL std::cout<<std::endl;
#define FLUSH std::cout<<std::flush;
#define SHOW(dist) {std::cerr << #dist << "\t:" << (dist) << "\n";}
#define SHOWVECTOR(v) {std::cerr << #v << "\t:";for(const auto& xxx : v){std::cerr << xxx << " ";}std::cerr << "\n";}
#define SHOWVECTOR2(v) {std::cerr << #v << "\t:\n";for(const auto& xxx : v){for(const auto& yyy : xxx){std::cerr << yyy << " ";}std::cerr << "\n";}}
#define SHOWQUEUE(a) {auto tmp(a);std::cerr << #a << "\t:";while(!tmp.empty()){std::cerr << tmp.front() << " ";tmp.pop();}std::cerr << "\n";}
#define SHOWSTACK(a) {auto tmp(a);std::cerr << #a << "\t:";while(!tmp.empty()){std::cerr << tmp.top() << " ";tmp.pop();}std::cerr << "\n";}
// utility
#define ALL(a) (a).begin(),(a).end()
#define FOR(w, a, n) for(int w=(a);w<(n);++w)
#define RFOR(w, a, n) for(int w=(n)-1;w>=(a);--w)
#define REP(w, n) for(int w=0;w<int(n);++w)
#define RREP(w, n) for(int w=int(n)-1;w>=0;--w)
#define IN(a, x, b) (a<=x && x<b)
template<class T> inline T CHMAX(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T CHMIN(T& a, const T b) { return a = (a > b) ? b : a; }
// test
template<class T> using V = std::vector<T>;
template<class T> using VV = V<V<T>>;

template<typename S, typename T>
std::ostream& operator<<(std::ostream& os, std::pair<S, T> p) {
	os << "(" << p.first << ", " << p.second << ")"; return os;
}

// type/const
#define int ll
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using PAIR = std::pair<int, int>;
using PAIRLL = std::pair<ll, ll>;
constexpr int INFINT = (1 << 30) - 1;                    // 1.07x10^ 9
constexpr int INFINT_LIM = (1LL << 31) - 1;              // 2.15x10^ 9
constexpr ll INFLL = 1LL << 60;                          // 1.15x10^18
constexpr ll INFLL_LIM = (1LL << 62) - 1 + (1LL << 62);  // 9.22x10^18
constexpr double EPS = 1e-10;
constexpr int MOD = 1000000007;
constexpr double PI = 3.141592653589793238462643383279;
template<class T, size_t N> void FILL(T(&a)[N], const T & val) { for (auto& x : a) x = val; }
template<class ARY, size_t N, size_t M, class T> void FILL(ARY(&a)[N][M], const T & val) { for (auto& b : a) FILL(b, val); }
template<class T> void FILL(std::vector<T> & a, const T & val) { for (auto& x : a) x = val; }
template<class ARY, class T> void FILL(std::vector<std::vector<ARY>> & a, const T & val) { for (auto& b : a) FILL(b, val); }
// ------------>8------------------------------------->8------------

// write [ LCA lca(g, root); ] when using this snippet.
class LCA {
private:
	const std::vector<std::vector<int>>& graph; // graph's list expression
	int root;
	int n; // the number of nodes
	int log2n; // = floor(log2(n)) + 1
	std::vector<std::vector<int>> parent; // parent[x][v] = a parent(above 2^x) of v (nonexistence -> -1)
	std::vector<int> depth; // the depth of each node

public:
	LCA(const std::vector<std::vector<int>>& graph, int root) :
		graph(graph), root(root), n(graph.size()),
		log2n(std::floor(std::log2(n) + 1)),
		parent(log2n, std::vector<int>(n, 0)), depth(n, 0)
	{
		init();
	}

	// Check the depth of each node(node "v" -> parent is "p", depth is "d")
	void dfs(int v, int p, int d) {
		std::stack<int> stack;
		stack.push(v);
		parent[0][v] = p;
		depth[v] = d;
		while (!stack.empty()) {
			int now = stack.top(); stack.pop();
			for (int i = 0; i < graph[now].size(); ++i) {
				int to = graph[now][i];
				if (to == parent[0][now]) continue;
				parent[0][to] = now;
				depth[to] = depth[now] + 1;
				stack.push(to); // Check each child of v
			}
		}
	}

	// Initialize
	void init() {
		// Initialize "parent[0]" and "depth"
		dfs(root, -1, 0);

		// Initialize "parent"
		for (int k = 0; k < log2n - 1; ++k) {
			for (int v = 0; v < n; ++v) {
				if (parent[k][v] < 0) { // If parent above 2^k of v is nonexistence
					parent[k + 1][v] = -1;
				}
				else {
					parent[k + 1][v] = parent[k][parent[k][v]];
				}
			}
		}
	}

	// Find LCA of (u, v)
	int lca(int u, int v) {
		// go up parent while depth of u and v is same
		if (depth[u] > depth[v]) std::swap(u, v);
		for (int k = 0; k < log2n; ++k) {
			if ((depth[v] - depth[u]) >> k & 1) {
				v = parent[k][v]; // go up to 2^k if k-th binary is 1
			}
		}

		if (u == v) return u; // this case is that v is in u's subtree

		// Find LCA by binary searching
		for (int k = log2n - 1; k >= 0; --k) {
			if (parent[k][u] != parent[k][v]) {
				u = parent[k][u];
				v = parent[k][v];
			}
		}
		return parent[0][u];
	}
};

signed main() {
	INIT;

	VAR(int, n);
	VEC_ROW(int, n - 1, a, b, c);
	std::vector<std::vector<PAIR>> g(n);
	std::vector<std::vector<int>> g_(n);
	REP(i, n-1) {
		g[a[i]].emplace_back(b[i], c[i]);
		g[b[i]].emplace_back(a[i], c[i]);
		g_[a[i]].emplace_back(b[i]);
		g_[b[i]].emplace_back(a[i]);
	}

	V<int> dist(n, INFLL);
	dist[0] = 0;
	{
		auto rec = [&](auto && f, int v, int par) -> void {
			for (auto& e : g[v]) if (e.first != par) {
				dist[e.first] = dist[v] + e.second;
				f(f, e.first, v);
			}
		};
		rec(rec, 0, -1);
	}

	VAR(int, Q);
	LCA lca(g_, 0);
	REP(_, Q) {
		VAR(int, x, y, z);
		int ans = INFLL;
		REP(i1, 3) {
			int w = lca.lca(x, y);
			int v = lca.lca(w, z);

			CHMIN(ans, dist[x] + dist[y] - dist[w] + dist[z] - 2 * dist[v]);

			int t;
			t = x;
			x = y;
			y = z;
			z = t;
		}
		OUT(ans)BR;
	}

	return 0;
}
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結果

ソースコード
