poj 2826 An Easy Problem?!(线段相交、模拟)

题目链接：

http://poj.org/problem?id=2826

题目大意：

给两条线段，然后有雨水落下来，问这两条线段组成的容器里面最多放置水的面积。

思路：

如果线段不相交，或者有一条线段是与x轴平行的，那么肯定不能容下水。能盛下水的情况就是两线段相交，然后有2个点在交点上方。但是有一种情况是，从y轴往下看的时候，有一条线段完全覆盖住了另一条，这个时候也是没有水进入的。

搞清楚情况以后，就是模拟了。

代码：

#include
#include
#include
#include
#define eps 1e-8
using namespace std;
struct Point{
	double x,y;
	Point(double x=0,double y=0):x(x),y(y){}
};
struct line{
	Point a,b;
	line(Point a=0,Point b=0):a(a),b(b){}
};
Point p[10005],ch[10005];
 line Line[5005];
Point operator + (Point A ,Point B){
	return Point(A.x+B.x,A.y+B.y);
}
Point operator - (Point A,Point B){
	return Point(A.x-B.x,A.y-B.y);
}
Point operator * (Point A,double p){
	return Point(A.x*p,A.y*p);
}
bool operator  < (const Point &a,const Point &b){
	return a.xP.y){
					tt++;
					if(i==1||i==2){A=p[i];ff=1;}
					else B=p[i];
				}
		}
		if(tt<2){
            printf("0.00\n");
            continue;
		}

		if(p[1].x!=p[2].x&&p[3].x!=p[4].x){
			k1=(p[2].y-p[1].y)/(p[2].x-p[1].x);
			k2=(p[4].y-p[3].y)/(p[4].x-p[3].x);
		}
		else {
			if(p[1].x==p[2].x&&p[3].x==p[4].x)printf("0.00\n");
			else {
				if(!SegmentProperIntersection(p[1],p[2],p[3],p[4])&&!(P==p[1])&&!(P==p[2])&&!(P==p[3])&&!(P==p[4])){
					printf("0.00\n");
					continue;
				}
					if(p[1].x==p[2].x){
						if(A.y>B.y){
								Point pp=intersection(l1,line(B,Point(B.x-1,B.y)));
								S=(pp.y-P.y)*(B.x-pp.x)*0.5;
						}
						else {
							Point pp=intersection(l2,line(A,Point(A.x-1,A.y)));
							S=(pp.y-P.y)*(A.x-pp.x)*0.5;
						}
					}
					else if(p[3].x==p[4].x){
						if(B.y>A.y){
							Point pp=intersection(l2,line(A,Point(A.x-1,A.y)));
							S=(pp.y-P.y)*(A.x-pp.x)*0.5;
						}
						else {
								Point pp=intersection(l1,line(B,Point(B.x-1,B.y)));
								S=(pp.y-P.y)*(B.x-pp.x)*0.5;
						}
					}
					printf("%.2f\n",fabs(S)+eps);
			}
			continue;
		}

		if(SegmentProperIntersection(p[1],p[2],p[3],p[4])||(P==p[1]||P==p[2]||P==p[3]||P==p[4])){
	//		printf("**\n");
			int ff=0;
			for(i=1;i<=4;i++){
				if(p[i].y>P.y){
					tt++;
					if(i==1||i==2){A=p[i];ff=1;}
					else B=p[i];
				}
			}

			if(tt<2){
				printf("0.00\n");
				continue;
			}
			else {
				if(k1*k2<0){
					if(A.y0){
						//printf("#$%\n");
						if(k1>k2){
							if(A.y>B.y){
								Point tmp=intersection(l2,line(A,Point(A.x,A.y-1)));
								tmp.y-=1;
								if(SegmentProperIntersection(p[3],p[4],A,tmp)){
									Point pp=intersection(l1,line(B,Point(B.x-1,B.y)));
									S=(B.x-pp.x)*(B.y-P.y)*0.5;
									printf("%.2f\n",fabs(S)+eps);
								}
								else {
									printf("0.00\n");
									continue;
								}
							}
							else {
								Point pp=intersection(l2,line(A,Point(A.x-1,A.y)));
								S=(A.x-pp.x)*(A.y-P.y)*0.5;
								printf("%.2f\n",fabs(S)+eps);
							}
						}

						else {
							if(B.y>A.y){
								Point tmp=intersection(l1,line(B,Point(B.x,B.y-1)));
								tmp.y-=1;
								if(SegmentProperIntersection(p[1],p[2],B,tmp)){
									Point pp=intersection(l2,line(A,Point(A.x-1,A.y)));
									S=(A.x-pp.x)*(A.y-P.y)	*0.5;
									printf("%.2f\n",fabs(S)+eps);
								}
								else {
									printf("0.00\n");
									continue;
								}
							}
							else {

								Point pp=intersection(l1,line(B,Point(B.x-1,B.y)));
								S=(B.x-pp.x)*(B.y-P.y)*0.5;
								printf("%.2f\n",fabs(S)+eps);
							}
						}
					}
					else {

						if(k1B.y){
								Point tmp=intersection(l2,line(A,Point(A.x,A.y-1)));
								tmp.y-=1;
								if(SegmentProperIntersection(p[3],p[4],A,tmp)){
									Point pp=intersection(l1,line(B,Point(B.x-1,B.y)));
									S=(B.x-pp.x)*(B.y-P.y)*0.5;
									printf("%.2f\n",fabs(S)+eps);
								}
								else {
									printf("0.00\n");
									continue;
									}
								}
								else {
									Point pp=intersection(l2,line(A,Point(A.x-1,A.y)));
									S=(pp.x-A.x)*(A.y-P.y)*0.5;
									printf("%.2f\n",fabs(S)+eps);
								}
							}
							else {
							//	printf("#$\n");
								if(B.y>A.y){
									Point tmp=intersection(l1,line(B,Point(B.x,B.y-1)));
									tmp.y-=1;
									if(SegmentProperIntersection(p[1],p[2],B,tmp)){
									Point pp=intersection(l2,line(A,Point(A.x-1,A.y)));
									S=(A.x-pp.x)*(A.y-P.y)	*0.5;
									printf("%.2f\n",fabs(S)+eps);
									}
									else {
										printf("0.00\n");
										continue;
										}
								}
								else {
									Point pp=intersection(l1,line(B,Point(B.x-1,B.y)));
									S=(pp.x-B.x)*(B.y-P.y)*0.5;
									printf("%.2f\n",fabs(S)+eps);
								}
							}
					}
				}
			}
		}
		else {
			printf("0.00\n");
			continue;
		}
	}
}
/*
1 1 1 4
0 0 2 3

*/