HDU - 5033 Building（单调栈）

Building

Time Limit: 5000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

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 
     
     
     
     

 

题目大意：
在一个平地上右n个摩天轮，题目告诉你每个摩天轮的位置，和高度，现在马特站在q个位置上，问马特站在每个位置上，他能看到最大仰望天空的角度为多少？
1<=N<=10^5 1<=Q<=10^5

解析：
这题的数据量很大，如果直接暴力的话肯定超时。
所以要用单调栈进行优化，将人和楼组合在一起，按照位置从小到大排序后，维护一个从左到右凸的高度下降的单调栈，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，那么栈顶和人所构成的仰角就是正向答案。但是这只是正向的答案，所以反向扫描所有的位置，维护出一个从右到左凸的单调栈，求出反向的答案，和之前的结果相加就是最终答案。

总结：在做这题时把pi写成int类型了，白白浪费了一个下午的时间要好好吸取教训。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <stack>
#include <algorithm>
using namespace std;
const int N = 100005 * 2;
const double PI = acos(-1.0);
struct Node {
	double x,y;
	int flag;
}p[N],st[N];
double ans[N];
int n,q;
bool cmp(Node a,Node b) {
	return a.x < b.x;
}
double getAngle(Node a,Node b) {
	return atan(fabs(b.x - a.x) / a.y);
}
double rate(Node a,Node b) {
	return (a.y - b.y) / (a.x - b.x);
}
bool judge_front(Node a,Node b,Node c) {
	double k1 = rate(a,b);
	double k2 = rate(a,c);
	if(k1 < 0 && k2 < 0 && fabs(k1) >= fabs(k2)) {
		return true;
	}else {
		return false;
	}
}
void solve_front() {
	int top = 0;
	for(int i = 0; i < n+q; i++) {
		if(p[i].flag) {
			while(top >= 2 && judge_front(st[top-2],st[top-1],p[i]) ) {
				top--;
			}
			ans[p[i].flag] += getAngle(st[top-1],p[i]);
		}else {
			while(top && st[top-1].y <= p[i].y) {
				top--;
			}
			while(top >= 2 && judge_front(st[top-2],st[top-1],p[i])) {
				top--;
			}
			st[top] = p[i];
			top++;
		}
	}
}
bool judge_back(Node a,Node b,Node c) {
	double k1 = rate(a,b);
	double k2 = rate(a,c);
	if(k1 > 0 && k2 > 0 && fabs(k1) >= fabs(k2)) {
		return true;
	}else {
		return false;
	}
}
void solve_back() {
	int top = 0;
	for(int i = n+q-1; i >= 0; i--) {
		if(p[i].flag) {
			while(top >= 2 && judge_back(st[top-1],st[top-2],p[i]) ) {
				top--;
			}
			ans[p[i].flag] += getAngle(st[top-1],p[i]);
		}else {
			while(top && st[top-1].y <= p[i].y) {
				top--;
			}
			while(top >= 2 && judge_back(st[top-2],st[top-1],p[i])) {
				top--;
			}
			st[top++] = p[i];
		}
	}
}
int main() {
	int t , cas = 1;
	scanf("%d",&t);
	while(t--) {
		scanf("%d",&n);
		for(int i = 0; i < n; i++) {
			scanf("%lf%lf",&p[i].x,&p[i].y);
			p[i].flag = 0;
		}
		scanf("%d",&q);
		for(int i = n; i < n+q; i++) {
			scanf("%lf",&p[i].x);
			p[i].y = 0;
			p[i].flag = i - n + 1;
		}
		sort(p,p+n+q,cmp);
		memset(ans,0,sizeof(ans));
		solve_front();
		solve_back();
		printf("Case #%d:\n",cas++);
		for(int i = 1; i <= q; i++) {
			printf("%.10lf\n",ans[i] * 180 / PI);
		}
	}
	return 0;
}