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POJ2540-Hotter Colder(半平面交)

时间:2014-08-20 21:11:12      阅读:378      评论:0      收藏:0      [点我收藏+]

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Hotter Colder
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2523   Accepted: 1045

Description

The children‘s game Hotter Colder is played as follows. Player A leaves the room while player B hides an object somewhere in the room. Player A re-enters at position (0,0) and then visits various other positions about the room. When player A visits a new position, player B announces "Hotter" if this position is closer to the object than the previous position; player B announces "Colder" if it is farther and "Same" if it is the same distance.

Input

Input consists of up to 50 lines, each containing an x,y coordinate pair followed by "Hotter", "Colder", or "Same". Each pair represents a position within the room, which may be assumed to be a square with opposite corners at (0,0) and (10,10).

Output

For each line of input print a line giving the total area of the region in which the object may have been placed, to 2 decimal places. If there is no such region, output 0.00.

Sample Input

10.0 10.0 Colder
10.0 0.0 Hotter
0.0 0.0 Colder
10.0 10.0 Hotter

Sample Output

50.00
37.50
12.50
0.00

题意:在(0,0)到(10,10)的范围内,还在藏在某个点,另一个人从起点出发,每走到一个点,如果离藏匿点近了就是“Hotter”,如果远了就是“Colder”,否则就是“Same” 

对于走的每个点,输出可能藏匿的面积大小。

思路: 可以转化为  aX+bY+c 与0的大小比较,因此可以用半平面交求解,每次找出中点和中垂线即可。


#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define REP(_,a,b) for(int _ = (a); _ < (b); _++)
#define sz(s)  (int)((s).size())
typedef long long ll;
const double eps = 1e-9;
const int maxn = 200;
struct Point{
	double x,y;
	Point(double x=0.0,double y = 0.0):x(x),y(y){}
};
vector<Point> vP;
Point poly[maxn];
typedef Point Vector;
struct Line {
	Point P;
	Vector v;
	double ang;
	Line(){}
	Line(Point P,Vector v):P(P),v(v){
		ang = atan2(v.y,v.x);
	}
	bool operator <(const Line&L) const{
		return ang < L.ang;
	}
};
Line L[maxn];
int lcnt;
Vector operator + (Vector A,Vector B) {
	return Vector(A.x+B.x,A.y+B.y);
}
Vector operator - (Vector A,Vector B){
	return Vector(A.x-B.x,A.y-B.y);
}
Vector operator * (Vector A,double p){
	return Vector(A.x*p,A.y*p);
}
Vector operator / (Vector A,double p){
	return Vector(A.x/p,A.y/p);
}
bool operator < (const Point &a,const Point &b){
	return a.x < b.x || (a.x==a.y && a.y < b.y);
}
int dcmp(double x){
	if(fabs(x) < eps) return 0;
	else return x < 0? -1:1;
}
bool operator == (const Point &a,const Point &b){
	return dcmp(a.x-b.x)==0&& dcmp(a.y-b.y)==0;
}
double Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}
double Length(Vector A) {return sqrt(Dot(A,A));}
double Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}
double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}
Vector Rotate(Vector A,double rad) {return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); }
Vector Normal(Vector A) {
	double L = Length(A);
	return Vector(-A.y/L,A.x/L);
}
bool OnLeft(Line L,Point p){
	return Cross(L.v,p-L.P) > 0;
}
Point GetIntersection(Line a,Line b){
	Vector u = a.P-b.P;
	double t = Cross(b.v,u) / Cross(a.v,b.v);
	return a.P+a.v*t;
}
int HalfplaneIntersection(Line* L,int n,Point* poly){
	sort(L,L+n);
	int first,last;
	Point *p = new Point[n];
	Line *q = new Line[n];
	q[first=last=0] = L[0];
	for(int i = 1; i < n; i++){
		while(first < last && !OnLeft(L[i],p[last-1])) last--;
		while(first < last && !OnLeft(L[i],p[first])) first++;
		q[++last] = L[i];
		if(fabs(Cross(q[last].v,q[last-1].v))<eps) {
			last--;
			if(OnLeft(q[last],L[i].P)) q[last] = L[i];
		}
		if(first<last) p[last-1] = GetIntersection(q[last-1],q[last]);
	}
	while(first < last && !OnLeft(q[first],p[last-1])) last--;
	if(last - first <=1) return 0;
	p[last] = GetIntersection(q[last],q[first]);
	int m = 0;
	for(int i = first; i <= last; i++) poly[m++] = p[i];
	return m;
}
void init(){
	vP.clear();
	lcnt = 0;
	vP.push_back(Point(0,0));
	vP.push_back(Point(10,0));
	vP.push_back(Point(10,10));
	vP.push_back(Point(0,10));
	int n = sz(vP);
	#define next(i) ((i)+1)%n
	REP(_,0,n){
		L[_] = Line(vP[_],vP[next(_)]-vP[_]);
		lcnt++;
	}
}
int main(){
	
	double x,y;
	string st;
	init();
	Point pre = Point(0,0),cur;
	bool flag = true;
	while(cin >> x >> y >> st){
		if(!flag){
			printf("0.00\n");
			continue;
		}
		cur = Point(x,y);
		if(st[0]=='S'){
			printf("0.00\n");
			flag = false;
			continue;
		}else if(st[0]=='H'){
			L[lcnt++] = Line((pre+cur)/2,Normal(pre-cur)) ;
		}else{
			L[lcnt++] = Line((pre+cur)/2,Normal(cur-pre));	
		}
		pre = Point(x,y);
		int m = HalfplaneIntersection(L,lcnt,poly);
		if(m==0){
			printf("0.00\n");
			flag = false;
		}else{
			#define next(i) ((i)+1)%m
			double area = 0.0;
			REP(_,0,m) { 
				area += Cross(poly[_],poly[next(_)])/2;
			}
			printf("%.2f\n",fabs(area));	
		}
	}	
	return 0;
}


POJ2540-Hotter Colder(半平面交),布布扣,bubuko.com

POJ2540-Hotter Colder(半平面交)

标签:des   style   blog   color   os   io   strong   for   

原文地址:http://blog.csdn.net/mowayao/article/details/38711623

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