【HDU】5033 Building 单调栈

传送门：【HDU】5033 Building

题目分析：就单调栈用叉积左右维护一个上凸壳就好了。。。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define clr( a , x ) memset ( a , x , sizeof a )
#define CPY( a , x ) memcpy ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define mid ( ( l + r ) >> 1 )
#define root 1 , 1 , 100000
#define rt o , l , r

const int MAXN = 100005 ;
const double eps = 1e-8 ;
const double pi = acos ( -1.0 ) ;

struct Point {
	double x , y ;
	int idx ;
	Point () {}
	Point ( double x , double y ) : x ( x ) , y ( y ) {}
	bool operator < ( const Point& p ) const {
		return x < p.x ;
	}
	Point operator - ( const Point& p ) const {
		return Point ( x - p.x , y - p.y ) ;
	}
} S[MAXN << 1] , p[MAXN << 1] ;

int top , cnt ;
int n , q ;
double ans[MAXN] ;

int dcmp ( double x ) {
	return ( x > eps ) - ( x < -eps ) ;
}

int cross ( const Point& a , const Point& b ) {
	return dcmp ( a.x * b.y - a.y * b.x ) ;
}

void solve () {
	double x , y ;
	top = cnt = 0 ;
	scanf ( "%d" , &n ) ;
	rep ( i , 0 , n ) {
		scanf ( "%lf%lf" , &x , &y ) ;
		p[cnt] = Point ( x , y ) ;
		p[cnt].idx = -1 ;
		++ cnt ;
	}
	scanf ( "%d" , &q ) ;
	rep ( i , 0 , q ) {
		scanf ( "%lf" , &x ) ;
		p[cnt] = Point ( x , 0 ) ;
		p[cnt].idx = i ;
		++ cnt ;
	}
	sort ( p , p + cnt ) ;
	rep ( i , 0 , cnt ) {
		while ( top > 1 && cross ( S[top - 1] - p[i] , S[top - 2] - p[i] ) <= 0 ) -- top ;
		if ( ~p[i].idx ) {
			double x = p[i].x - S[top - 1].x ;
			double y = S[top - 1].y ;
			ans[p[i].idx] = atan ( x / y ) * 180.0 / pi ;
		} else S[top ++] = p[i] ;
	}
	top = 0 ;
	rev ( i , cnt - 1 , 0 ) {
		while ( top > 1 && cross ( S[top - 1] - p[i] , S[top - 2] - p[i] ) >= 0 ) -- top ;
		if ( ~p[i].idx ) {
			double x = S[top - 1].x - p[i].x ;
			double y = S[top - 1].y ;
			ans[p[i].idx] += atan ( x / y ) * 180.0 / pi ;
		} else S[top ++] = p[i] ;
	}
	rep ( i , 0 , q ) printf ( "%.10f\n" , ans[i] ) ;
}

int main () {
	int T , cas = 0 ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) {
		printf ( "Case #%d:\n" , ++ cas ) ;
		solve () ;
	}
	return 0 ;
}