[hdu5033]单调队列

题意：x轴上有n棵树，询问你站在某个点的视角。从左至右，单调队列（类似凸包）维护下。我强迫症地写了个模板QAQ

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <cstdlib>
  5 #include <cstring>
  6 #include <map>
  7 #include <queue>
  8 #include <deque>
  9 #include <cmath>
 10 #include <vector>
 11 #include <ctime>
 12 #include <cctype>
 13 #include <set>
 14
 15 using namespace std;
 16
 17 #define mem0(a) memset(a, 0, sizeof(a))
 18 #define lson l, m, rt << 1
 19 #define rson m + 1, r, rt << 1 | 1
 20 #define define_m int m = (l + r) >> 1
 21 #define Rep(a, b) for(int a = 0; a < b; a++)
 22 #define lowbit(x) ((x) & (-(x)))
 23 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 24 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 25 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 26
 27 typedef double db;
 28 typedef long long LL;
 29 typedef pair<int, int> pii;
 30 typedef multiset<int> msi;
 31 typedef multiset<int>::iterator msii;
 32 typedef set<int> si;
 33 typedef set<int>::iterator sii;
 34 typedef vector<int> vi;
 35
 36 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
 37 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1};
 38 const int maxn = 1e5 + 7;
 39 const int maxm = 1e5 + 7;
 40 const int maxv = 1e7 + 7;
 41 const int MD = 1e9 +7;
 42 const int INF = 1e9 + 7;
 43 const double PI = acos(-1.0);
 44 const double eps = 1e-10;
 45
 46 struct Point {
 47     int x, y;
 48     bool operator < (const Point &opt) const {
 49         return x < opt.x || x == opt.x && y < opt.y;
 50     }
 51     double abs() {
 52         return sqrt((double)x * x + (double)y * y);
 53     }
 54     Point operator - (const Point &opt) const {
 55         return Point(x - opt.x, y - opt.y);
 56     }
 57     double operator * (const Point &opt) const {
 58         return (double)x * opt.x + (double)y * opt.y;
 59     }
 60     constructInt2(Point, x, y);
 61     void inp() {
 62         scanf("%d %d", &x, &y);
 63     }
 64     void outp() {
 65         printf("(%d, %d), ", x, y);
 66     }
 67 };
 68
 69 bool cmp(const pii &a, const pii &b) {
 70     return a.first < b.first;
 71 }
 72
 73 double Cross(const Point &a, const Point &b) {
 74     return (double)a.x * b.y - (double)a.y * b.x;
 75 }
 76
 77 template<class T> struct MonotoneQueue{
 78     deque<T> Q;
 79     MonotoneQueue<T>() { Q.clear(); }
 80     T back() { return Q.back(); }
 81     T back2() { if (Q.size() < 2) return T(); return *(Q.end() - 2); }
 82     T front() { return Q.front(); }
 83     void clear() { Q.clear(); }
 84     bool empty() { return Q.empty(); }
 85     void add_back(T x) { while (!empty() && !(back() - back2() < x - back2())) Q.pop_back(); Q.push_back(x); }
 86     void pop_front() { Q.pop_front(); }
 87 };
 88
 89 double toDegree(double x) { return x * 180.0 / PI; }
 90
 91 Point p[maxn], L[maxn], R[maxn];
 92 pii qry[maxn];
 93 int n, m;
 94
 95 struct Node {
 96     Point dot;
 97     bool operator < (const Node &a) const {
 98         return Cross(dot, a.dot) < eps;
 99     }
100     Node operator - (const Node &opt) const {
101         return Node(dot - opt.dot);
102     }
103     Node(Point dot = Point()): dot(dot) {}
104 };
105 MonotoneQueue<Node> DQ;
106
107 void work(Point a[]) {
108     DQ.clear();
109     DQ.add_back(p[0]);
110     int now = 1;
111     for (int i = 0; i < m; i++) {
112         while (now < n && p[now].x < qry[i].first) {
113             DQ.add_back(Node(p[now++]));
114         }
115         DQ.add_back(Node(Point(qry[i].first, 0)));
116         a[qry[i].second] = (DQ.back2() - DQ.back()).dot;
117     }
118 }
119
120 int main() {
121     //freopen("in.txt", "r", stdin);
122     int T, cas = 0;
123     cin >> T;
124     while (T--) {
125         cin >> n;
126         for (int i = 0; i < n; i++) {
127             p[i].inp();
128         }
129         sort(p, p + n);
130         cin >> m;
131         for (int i = 0, x; i < m; i++) {
132             scanf("%d", &x);
133             qry[i] = make_pair(x, i);
134         }
135         sort(qry, qry + m, cmp);
136
137         work(L);
138
139         for (int i = 0; i < n; i++) {
140             p[i].x = maxv - p[i].x;
141         }
142         sort(p, p + n);
143         for (int i = 0; i < m; i++) {
144             qry[i].first = maxv - qry[i].first;
145         }
146         sort(qry, qry + m, cmp);
147
148         work(R);
149         for (int i = 0; i < m; i++) {
150             R[i].x = -R[i].x;
151         }
152
153         printf("Case #%d:\n", ++cas);
154         for (int i = 0; i < m; i++) {
155             printf("%.10f\n", toDegree(acos(L[i] * R[i] / L[i].abs() / R[i].abs())));
156         }
157     }
158     return 0;
159 }