結果
問題 | No.898 tri-βutree |
ユーザー |
|
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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 21 |
ソースコード
// 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; }// testtemplate<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 llusing 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^ 9constexpr int INFINT_LIM = (1LL << 31) - 1; // 2.15x10^ 9constexpr ll INFLL = 1LL << 60; // 1.15x10^18constexpr ll INFLL_LIM = (1LL << 62) - 1 + (1LL << 62); // 9.22x10^18constexpr 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 expressionint root;int n; // the number of nodesint log2n; // = floor(log2(n)) + 1std::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 nodepublic: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}}}// Initializevoid 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 nonexistenceparent[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 sameif (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 searchingfor (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;}