제출 #1129214

#제출 시각아이디문제언어결과실행 시간메모리
1129214RedGrey1993사다리꼴 (balkan11_trapezoid)C++20
43 / 100
344 ms49316 KiB
#include <bits/stdc++.h> using namespace std; template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &pa) { is >> pa.first >> pa.second; return is; } template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; } template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << "(" << pa.first << "," << pa.second << ")"; return os; } template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; } template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; } template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template <typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; } template <typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp) { os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; } template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; } template <typename T> void resize_array(vector<T> &vec, int len) { vec.resize(len); } template <typename T, typename... Args> void resize_array(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) resize_array(v, args...); } template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); } template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); } long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; } mt19937 mrand(random_device{}()); int rnd(int x) { return mrand() % x; } // 函数名后面加 ll 就可以计算 long long类型对应的结果 // __builtin_ffs(x) // 返回x中最后一个为1的位是从后向前的第几位 // __builtin_popcount(x) // x中1的个数 // __builtin_ctz(x) // x末尾0的个数。x=0时结果未定义。 // __builtin_clz(x) // x前导0的个数。x=0时结果未定义。 // __builtin_parity(x) // x中1的奇偶性。 #define highest_bit1_index(x) (31 - __builtin_clz(x)) #define highest_bit1_index_ll(x) (63 - __builtin_clzll(x)) #define rep(i, a, n) for (int i = a; i < (n); i++) #define per(i, a, n) for (int i = (n)-1; i >= a; i--) #define pb push_back #define mp make_pair #define all(x) (x).begin(), (x).end() #define fi first #define se second #define sz(x) ((int)(x).size()) typedef vector<int> vi; typedef long long ll; typedef unsigned int uint; typedef unsigned long long ull; typedef pair<int, int> pii; typedef double db; #if DEBUG #define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl; #else #define dbg(x) #endif // For example: SegmentTree<int> st(n, [](int a, int b) { return max(a, b); }); template <typename T> class SegmentTree { public: SegmentTree(int node_num, function<T(T, T)> f, T default_v = T()) : f_(f), default_value_(default_v) { node_num_ = node_num; offset_ = 1; while (offset_ < node_num) offset_ *= 2; nodes_.resize(offset_ * 2, default_v); } friend istream& operator>>(istream& in, SegmentTree<T>& st) { for (int i = 0; i < st.node_num_; ++i) { cin >> st.nodes_[i + st.offset_ - 1]; } return in; } // Init 前需要先赋初始值给nodes_ // rep(i,0,n) { // nodes_[i + offset_ - 1] = A[i]; // } void Init() { for (int k = offset_ - 2; k >= 0; --k) { nodes_[k] = f_(nodes_[k * 2 + 1], nodes_[k * 2 + 2]); } } // 将第k个值更新为a void Update(int k, T a) { // 叶子节点 k += offset_ - 1; nodes_[k] = a; // 向上更新 while (k > 0) { k = (k - 1) / 2; nodes_[k] = f_(nodes_[k * 2 + 1], nodes_[k * 2 + 2]); } } // 迭代写法,效率更高(多找几道题验证下) // T Query(int lo, int hi) { // T ra = default_value_, rb = default_value_; // for (lo += offset_ - 1, hi += offset_ - 1; lo < hi; lo /= 2, hi /= 2) { // if (!(lo & 1)) ra = f_(ra, nodes_[lo++]); // if (!(hi & 1)) rb = f_(rb, nodes_[--hi]); // } // return f_(ra, rb); // } // 递归写法,更好理解 T Query(int a, int b) { return Query(a, b, 0, 0, offset_); } private: // 求[a,b)区间的目标值 // 后面的参数是为了计算方便传入的 // k是节点的编号,l,r表示这个节点代表区间[l,r) // 外部调用时,使用query(a,b,0,0,n); T Query(int a, int b, int k, int l, int r) { // 如果[a,b)和[l,r)不相交,返回0 if (r <= a || b <= l) return default_value_; // 如果[a,b)完全包含[l,r),返回当前节点值 if (a <= l && r <= b) return nodes_[k]; else { T v1 = Query(a, b, k * 2 + 1, l, (l + r) / 2); T v2 = Query(a, b, k * 2 + 2, (l + r) / 2, r); return f_(v1, v2); } } int node_num_; int offset_; vector<T> nodes_; function<T(T, T)> f_; T default_value_; }; constexpr int MOD = 30013; class Solution { struct Line { int up,down; bool is_left; int id; pii max_independent_set = mp(0, 0); friend ostream& operator<<(ostream& os, const Line& line) { os << "[" << (line.is_left ? "L" : "R") << "](" << line.up << "," << line.down << ")"; return os; } bool operator<(const Line& other) const { return down == other.down ? up < other.up : down < other.down; } }; public: void Solve() { int n; while(cin>>n) { // 存储梯形的坐标信息 // 每个梯形由两个点对表示:{{左上,左下}, {右上,右下}} vector<Line> trapezoids; set<int> coordinates; // 读入n个梯形的坐标 for (int i = 0; i < n; i++) { int upper_left, upper_right; // 上边两端点的x坐标 int bottom_left, bottom_right; // 下边两端点的x坐标 cin >> upper_left >> upper_right >> bottom_left >> bottom_right; coordinates.insert(upper_left); coordinates.insert(upper_right); coordinates.insert(bottom_left); coordinates.insert(bottom_right); trapezoids.push_back({upper_left, bottom_left, true, i}); trapezoids.push_back({upper_right, bottom_right, false, i}); } map<int, int> cordinate_compress; int i = 0; for (auto x : coordinates) { cordinate_compress[x] = i++; } sort(trapezoids.begin(), trapezoids.end()); map<int, int> right_line_id_to_index; for(int i = 0; i < trapezoids.size(); i++) { if(!trapezoids[i].is_left) { right_line_id_to_index[trapezoids[i].id] = i; } } SegmentTree<pii> st(coordinates.size(), [](pii a, pii b) { if (a.first > b.first) return a; else if (a.first < b.first) return b; else return mp(a.first, (a.second + b.second) % MOD); }, mp(0, 0)); pii ans = mp(0, 0); for(int i = 0; i < trapezoids.size(); i++) { if(trapezoids[i].is_left) { pii ret = st.Query(0, cordinate_compress[trapezoids[i].up]); if (ret.second == 0) ret.second = 1; ret.first = ret.first + 1; auto& right_line = trapezoids[right_line_id_to_index[trapezoids[i].id]]; right_line.max_independent_set = ret; ans = max(ans, ret); } else { auto& right_line = trapezoids[i]; st.Update(cordinate_compress[right_line.up], right_line.max_independent_set); } } // 输出所有梯形的坐标信息 cout << ans.first << " " << ans.second << endl; } cout.flush(); } private: }; // # define FILE_IO 1 void set_io(const string &name = "") { ios::sync_with_stdio(false); cin.tie(nullptr); #if FILE_IO if (!name.empty()) { freopen((name + ".in").c_str(), "r", stdin); freopen((name + ".out").c_str(), "w", stdout); } #endif } int main() { set_io("tmp"); Solution().Solve(); return 0; }
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