分析：
利用单调栈分别算出最右的能够看见的楼，和最左的能够看见的楼，然后计算。
1 #include <iostream> 2 #include <algorithm> 3 #include <vector> 4 #include <map> 5 #include <cstring> 6 #include <set> 7 #include <bitset> 8 #include <cstdio> 9 #include <cmath> 10 11 #define esp 1e-8 12 #pragma comment(linker, "/STACK:102400000,102400000") 13 #define in freopen("solve_in.txt", "r", stdin); 14 #define pf(x) ((x)*(x)) 15 #define bug(x) printf("Line %d: >>>>>> ", (x)); 16 #define lson l, m, rt<<1 17 #define rson m+1, r, rt<<1|1 18 #define pb push_back 19 #define mp make_pair 20 #define pi acos(-1.0) 21 22 typedef unsigned short US; 23 24 25 using namespace std; 26 27 const int maxn = (int)1e5 + 100; 28 int sta[maxn*2]; 29 int dblcmp(double x) { 30 if(fabs(x) < esp) return 0; 31 return x > 0 ? 1 : -1; 32 } 33 struct Node { 34 double x, y; 35 int type, id; 36 bool operator < (const Node &o)const { 37 return dblcmp(x-o.x) < 0; 38 } 39 } p[maxn*2]; 40 41 double cal(int x, int y) { 42 return (p[y].y-p[x].y)/(p[y].x-p[x].x); 43 } 44 int l[maxn], r[maxn]; 45 double ans[maxn]; 46 double x[maxn]; 47 48 int main() { 49 50 int T; 51 int n, q; 52 for(int t = scanf("%d", &T); t <= T; t++) { 53 printf("Case #%d: ", t); 54 int top = 0; 55 scanf("%d", &n); 56 int cnt = 0; 57 for(int i = 1; i <= n; i++) { 58 scanf("%lf%lf", &p[cnt].x, &p[cnt].y); 59 p[cnt++].type = 0; 60 } 61 scanf("%d", &q); 62 for(int i = 1; i <= q; i++) { 63 scanf("%lf", &p[cnt].x); 64 p[cnt].y = 0.0; 65 p[cnt].id = i; 66 x[i] = p[cnt].x; 67 p[cnt++].type = 1; 68 } 69 sort(p, p+cnt); 70 top = 0; 71 for(int i = cnt-1; i >= 0; i--) { 72 while(top >= 2 && cal(i, sta[top]) < cal(sta[top], sta[top-1])) 73 top--; 74 if(p[i].type){ 75 int id = p[i].id; 76 r[id] = sta[top]; 77 } 78 sta[++top] = i; 79 } 80 top = 0; 81 for(int i = 0; i < cnt; i++){ 82 while(top >= 2 && cal(i, sta[top]) > cal(sta[top], sta[top-1])) 83 top--; 84 if(p[i].type){ 85 int id = p[i].id; 86 l[id] = sta[top]; 87 } 88 sta[++top] = i; 89 } 90 for(int i = 1; i <= q; i++){ 91 double d1 = sqrt(pf(p[l[i]].y) + pf(p[l[i]].x-x[i])); 92 double d2 = sqrt(pf(p[r[i]].y) + pf(p[r[i]].x-x[i])); 93 double tmp = acos(fabs(p[l[i]].x-x[i])/d1) + acos(fabs(p[r[i]].x-x[i])/d2); 94 tmp = (pi-tmp)/pi*180.0; 95 printf("%.10f ", tmp); 96 } 97 } 98 return 0; 99 }