POJ 1755

列出不等式后,把同时除Z把它去掉。

注意了,这里应该 是把直线变两点表示的向量更为简单,我开始就直接用直线写,后来,唉,写不下去了。。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;
const int MAX=110;
const double eps=1e-10;
const double inf=1e10;

struct point{
	double x,y;
};
point operator -(point x,point y){
	point t;
	t.x=x.x-y.x; t.y=x.y-y.y;
	return t;
}

double operator *(point x,point y){
	return x.x*y.y-x.y*y.x;
}

struct line{
	point a,b;
	double angle;
};
int n,lnum;
double A[MAX],B[MAX],C[MAX];
line le[MAX],st[MAX];

int DB(double d){
	if(fabs(d)<eps) return 0;
	if(d>0) return 1;
	return -1;
}

void addLine(double x1,double y1,double x2,double y2){
	point a; a.x=x1; a.y=y1;
	point b; b.x=x2; b.y=y2;
	le[lnum].a=a; le[lnum].b=b;
	le[lnum].angle=atan2(y2-y1,x2-x1);
}

bool cmp( line x, line y){
	if(fabs(x.angle-y.angle)<eps){
		if((y.b-x.a)*(x.b-y.a)>eps)
		return true;
		return false;
	}
	return x.angle<y.angle;	
}

void getIntersect(line l1, line l2, point& p) {  
    double dot1,dot2;  
    dot1 = (l1.b-l2.a)*(l1.a-l2.a);  
    dot2 = (l2.b-l1.b)*( l1.a-l1.b);  
    p.x = (l2.a.x * dot2 + l2.b.x * dot1) / (dot2 + dot1);  
    p.y = (l2.a.y * dot2 + l2.b.y * dot1) / (dot2 + dot1);  
}  
  
  
bool judge(line l0, line l1, line l2) {  
    point p;  
    getIntersect(l1, l2, p);  
    return DB((l0.a-p)*(l0.b-p)) <=0;  
 }


bool pare(line x,line y){
	return fabs((x.b-x.a)*(y.b-y.a))<eps;
}

bool half(){
	int top=1,bot=0;
	sort(le,le+lnum,cmp);
	int tmp=1;
	for(int i=1;i<lnum;i++){
		if(fabs(le[i].angle-le[tmp-1].angle)>eps) le[tmp++]=le[i];
	}
	lnum=tmp;
	st[0]=le[0]; st[1]=le[1];
	for(int i=2;i<lnum;i++){
		while(bot<top&&judge(le[i], st[top-1], st[top])) top--;
		while(bot<top&&judge(le[i],st[bot+1],st[bot])) bot++;
		st[++top]=le[i];
	}
	while(bot<top&&judge(st[bot],st[top-1],st[top])) top--;
	while(bot<top&&judge(st[top],st[bot+1],st[bot])) bot++;
	if(top<=bot+1) return false;
	return true;
}

bool slove(int s){
	int i, j, k;
    double x1, y1, x2, y2, a, b, c;  
	lnum=0; 
	addLine(0, 0, inf, 0 );   lnum++;
    addLine(inf, 0, inf, inf );  lnum++;
    addLine(inf, inf, 0, inf );  lnum++;
    addLine(0, inf, 0, 0 );  lnum++;   
    for (j = 0; j < n; j++)  
    	if (s != j) {  
			a = 1.0 / A[j] - 1.0 / A[s];  
   			b = 1.0 / B[j] - 1.0 / B[s];  
      		c = 1.0 / C[j] - 1.0 / C[s];  
        	int d1 = DB(a);  
         	int d2 = DB(b);  
          	int d3 = DB(c);   
           	if (!d1) {  
            	if (!d2) {  
             		if (d3 <= 0) {  
               			return false;  
                  	}   
                  	continue;
               	}  
                x1 = 0;  
                x2 = d2;  
                y1 = y2 = - c / b;  
             }  
             else {  
             	if (!d2) {  
              		x1 = x2 = - c / a;  
                	y1 = 0;  
                    y2 = -d1;  
                 }  
                 else {  
                 	x1 = 0; y1 = - c / b;  
                  	x2 = d2;  
                   	y2 = -(c + a * x2) / b;  
                 }  
             }  
             addLine(x1, y1, x2, y2 );
             lnum++;
	}  
	if(!half()) return false;
	return true;
}

int main(){
	while(scanf("%d",&n)!=EOF){
		for(int i=0;i<n;i++)
		scanf("%lf%lf%lf",&A[i],&B[i],&C[i]);
		for(int i=0;i<n;i++){
			if(slove(i)) printf("Yes\n");
			else printf("No\n");
		}
	}
	return 0;
}

  

posted @ 2014-08-08 11:33  chenjunjie1994  阅读(246)  评论(0编辑  收藏  举报