hdu 5033 单调栈 ****

看出来是单调栈维护斜率，但是不会写，2333，原来是和询问放在一起的

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <cmath>
 7 typedef __int64 ll;
 8 using namespace std;
 9
10 const double PI = acos(-1.0);
11 const int maxn = 100005;
12
13 struct Node {
14     int x, h;
15     bool operator <(const Node &a)const {
16         return x < a.x;
17     }
18 } node[maxn<<2], stk[maxn<<2];
19 double ans[maxn];
20 int n, q;
21
22 int check(Node &a, Node &b, Node c) {
23     if (c.h <= 0)
24         c.h = 0;
25     return (ll)(a.h - c.h) * (c.x - b.x) >= (ll)(b.h - c.h) * (c.x - a.x);
26 }
27
28 double getAngle(Node a, Node b) {
29     return atan((double)(b.x-a.x)/(double)(a.h));
30 }
31
32 void cal() {
33     int head = 0;
34     for (int i = 0; i < n+q; i++) {
35         if (node[i].h <= 0) {
36             while (head >= 2 && check(stk[head-2], stk[head-1], node[i]))
37                 head--;
38             ans[-node[i].h] += getAngle(stk[head-1], node[i]);
39         }
40         else {
41             while (head && stk[head-1].h <= node[i].h)
42                 head--;
43             while (head >= 2 && check(stk[head-2], stk[head-1], node[i]))
44                 head--;
45             stk[head++] = node[i];
46         }
47     }
48 }
49
50 int main() {
51     int t, cas = 1;
52     scanf("%d", &t);
53     while (t--) {
54         scanf("%d", &n);
55         for (int i = 0; i < n; i++)
56             scanf("%d%d", &node[i].x, &node[i].h);
57         scanf("%d", &q);
58         for (int i = 0; i < q; i++) {
59             scanf("%d", &node[i+n].x);
60             node[i+n].h = -i;
61         }
62
63         memset(ans, 0, sizeof(ans));
64         sort(node, node+n+q);
65
66         cal();
67
68         reverse(node, node+n+q);
69         for (int i = 0; i < n+q; i++)
70             node[i].x = 10000000 - node[i].x;
71
72         cal();
73
74         printf("Case #%d:\n", cas++);
75         for (int i = 0; i < q; i++)
76             printf("%.10lf\n", ans[i] * 180.0 / PI);
77     }
78     return 0;
79 }