[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

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define Rep(a, b) for(int a = 0; a < b; a++)
 #define lowbit(x) ((x) & (-(x)))
 #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) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

 typedef double db;
 typedef long long LL;
 typedef pair<int, int> pii;
 typedef multiset<int> msi;
 typedef multiset<int>::iterator msii;
 typedef set<int> si;
 typedef set<int>::iterator sii;
 typedef vector<int> vi;

 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {, -, , , -, , , -};
 const int maxn = 1e5 + ;
 const int maxm = 1e5 + ;
 const int maxv = 1e7 + ;
 const int MD = 1e9 +;
 const int INF = 1e9 + ;
 const double PI = acos(-1.0);
 const double eps = 1e-;

 struct Point {
     int x, y;
     bool operator < (const Point &opt) const {
         return x < opt.x || x == opt.x && y < opt.y;
     }
     double abs() {
         return sqrt((double)x * x + (double)y * y);
     }
     Point operator - (const Point &opt) const {
         return Point(x - opt.x, y - opt.y);
     }
     double operator * (const Point &opt) const {
         return (double)x * opt.x + (double)y * opt.y;
     }
     constructInt2(Point, x, y);
     void inp() {
         scanf("%d %d", &x, &y);
     }
     void outp() {
         printf("(%d, %d), ", x, y);
     }
 };

 bool cmp(const pii &a, const pii &b) {
     return a.first < b.first;
 }

 double Cross(const Point &a, const Point &b) {
     return (double)a.x * b.y - (double)a.y * b.x;
 }

 template<class T> struct MonotoneQueue{
     deque<T> Q;
     MonotoneQueue<T>() { Q.clear(); }
     T back() { return Q.back(); }
     T back2() { if (Q.size() < ) return T(); return *(Q.end() - ); }
     T front() { return Q.front(); }
     void clear() { Q.clear(); }
     bool empty() { return Q.empty(); }
     void add_back(T x) { while (!empty() && !(back() - back2() < x - back2())) Q.pop_back(); Q.push_back(x); }
     void pop_front() { Q.pop_front(); }
 };

 double toDegree(double x) { return x * 180.0 / PI; }

 Point p[maxn], L[maxn], R[maxn];
 pii qry[maxn];
 int n, m;

 struct Node {
     Point dot;
     bool operator < (const Node &a) const {
         return Cross(dot, a.dot) < eps;
     }
     Node operator - (const Node &opt) const {
         return Node(dot - opt.dot);
     }
     Node(Point dot = Point()): dot(dot) {}
 };
 MonotoneQueue<Node> DQ;

 void work(Point a[]) {
     DQ.clear();
     DQ.add_back(p[]);
     int now = ;
     for (int i = ; i < m; i++) {
         while (now < n && p[now].x < qry[i].first) {
             DQ.add_back(Node(p[now++]));
         }
         DQ.add_back(Node(Point(qry[i].first, )));
         a[qry[i].second] = (DQ.back2() - DQ.back()).dot;
     }
 }

 int main() {
     //freopen("in.txt", "r", stdin);
     int T, cas = ;
     cin >> T;
     while (T--) {
         cin >> n;
         for (int i = ; i < n; i++) {
             p[i].inp();
         }
         sort(p, p + n);
         cin >> m;
         for (int i = , x; i < m; i++) {
             scanf("%d", &x);
             qry[i] = make_pair(x, i);
         }
         sort(qry, qry + m, cmp);

         work(L);

         for (int i = ; i < n; i++) {
             p[i].x = maxv - p[i].x;
         }
         sort(p, p + n);
         for (int i = ; i < m; i++) {
             qry[i].first = maxv - qry[i].first;
         }
         sort(qry, qry + m, cmp);

         work(R);
         for (int i = ; i < m; i++) {
             R[i].x = -R[i].x;
         }

         printf("Case #%d:\n", ++cas);
         for (int i = ; i < m; i++) {
             printf("%.10f\n", toDegree(acos(L[i] * R[i] / L[i].abs() / R[i].abs())));
         }
     }
     return ;
 }