Building
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1090 Accepted Submission(s): 309
Special Judge
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
比赛的时候没想到，看了别人的题解，http://blog.csdn.net/u011345136/article/details/39454537
方法太巧妙了，由于查询的位置没有建筑，把查询的位置和建筑物的位置放在一个数组里，把查询的位置特殊标记一下，取查询次序的相反值，这样不仅标记了查询的次序，又与建筑物分开了，最后从左往又扫一遍，取当前位置左边的斜率的最大值，从右往左扫一遍，取当前位置右边斜率的最大值。这样就可以算最大视角了。
代码：
#include
#include
#include
#include
#include
using namespace std; const int maxn=100000+100; const double pi = acos(-1.0); struct node { int x; int h; }a[maxn*2],stacks[maxn*2]; double ans[maxn]; int n,q; bool cmp(node a,node b) { return a.x
=(long long)(c.x-a.x)*(b.h-c.h); } double getangle(node a,node b) { return atan((double)(b.x-a.x)*1.0/(double)(a.h*1.0)); } void solve() { int head=0; for(int i=0;i
=2&&cross(stacks[head-2],stacks[head-1],a[i])) { head--; } ans[-a[i].h]+=getangle(stacks[head-1],a[i]); } else { while(head&&stacks[head-1].h<=a[i].h) { head--; } while(head>=2&&cross(stacks[head-2],stacks[head-1],a[i])) { head--; } stacks[head++]=a[i]; } } } int main() { int t; scanf("%d",&t); int ca=1; while(t--) { scanf("%d",&n); for(int i=0;i