Problem Description Once upon a time Matt went to a small town. The town was so small and narrow that he can regard the town as a pivot. There were some skyscrapers in the town, each located at position x
i with its height h
i. All skyscrapers located in different place. The skyscrapers had no width, to make it simple. As the skyscrapers were so high, Matt could hardly see the sky.Given the position Matt was at, he wanted to know how large the angle range was where he could see the sky. Assume that Matt's height is 0. It's guaranteed that for each query, there is at least one building on both Matt's left and right, and no building locate at his position.
Input The first line of the input contains an integer T, denoting the number of testcases. Then T test cases follow.
Each test case begins with a number N(1<=N<=10^5), the number of buildings.
In the following N lines, each line contains two numbers, x
i(1<=x
i<=10^7) and h
i(1<=h
i<=10^7).
After that, there's a number Q(1<=Q<=10^5) for the number of queries.
In the following Q lines, each line contains one number q
i, which is the position Matt was at.
Output For each test case, first output one line "Case #x:", where x is the case number (starting from 1).
Then for each query, you should output the angle range Matt could see the sky in degrees. The relative error of the answer should be no more than 10^(-4).
Sample Input
3
3
1 2
2 1
5 1
1
4
3
1 3
2 2
5 1
1
4
3
1 4
2 3
5 1
1
4
Sample Output
Case #1:
101.3099324740
Case #2:
90.0000000000
Case #3:
78.6900675260
题意：给你n栋楼，每栋楼有高度和坐标，现在询问你人站在某个位置，能看到的天空的角度
思路：将询问的人和楼组合在一起，排序后，维护一个凸的高度下降的单调栈(可以动手画一下)，然后每次查询到人的位置的时候，维护单调栈，使得图形是凸的，那么栈首和这个人就能构成答案了，还有是楼的时候也要维护这个栈，那么前后各遍历一次，就能通过正切值求解了
#include
#include
#include
#include
#include
#include
typedef __int64 ll; using namespace std; const double PI = acos(-1.0); const int maxn = 100005; struct Node { int x, h; bool operator <(const Node &a)const { return x < a.x; } } node[maxn<<2], stk[maxn<<2]; double ans[maxn]; int n, q; int check(Node &a, Node &b, Node c) { if (c.h <= 0) c.h = 0; return (ll)(a.h - c.h) * (c.x - b.x) >= (ll)(b.h - c.h) * (c.x - a.x); } double getAngle(Node a, Node b) { return atan((double)(b.x-a.x)/(double)(a.h)); } void cal() { int head = 0; for (int i = 0; i < n+q; i++) { if (node[i].h <= 0) { while (head >= 2 && check(stk[head-2], stk[head-1], node[i])) head--; ans[-node[i].h] += getAngle(stk[head-1], node[i]); } else { while (head && stk[head-1].h <= node[i].h) head--; while (head >= 2 && check(stk[head-2], stk[head-1], node[i])) head--; stk[head++] = node[i]; } } } int main() { int t, cas = 1; scanf("%d", &t); while (t--) { scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%d%d", &node[i].x, &node[i].h); scanf("%d", &q); for (int i = 0; i < q; i++) { scanf("%d", &node[i+n].x); node[i+n].h = -i; } memset(ans, 0, sizeof(ans)); sort(node, node+n+q); cal(); reverse(node, node+n+q); for (int i = 0; i < n+q; i++) node[i].x = 10000000 - node[i].x; cal(); printf("Case #%d:\n", cas++); for (int i = 0; i < q; i++) printf("%.10lf\n", ans[i] * 180.0 / PI); } return 0; }