poj2826

题意：平面上两条线段，问能接住多少雨水

方法：首先看两条线段能否相交，不相交肯定为0。

         如果相交，求得交点后判断4个端点中是否存在两个点其纵坐标大于交点坐标，如果不存在，肯定为0

         如果存在，判断高板是否将低板遮挡，过高于纵坐标的两个端点中较低的那个（p1)做垂线，如果交点存在且纵坐标大于p1纵坐标，

         则被遮挡，为0。否则过p1做一条水平线与另一条线段相交得交点p3则面积就是 fabs(cross(p,p1,p3))/2了

WA点：此题考察线段能接多少雨水而非能盛多少雨水。存在能盛雨水，但是雨水无法垂直落入的情况，比较难想。

 

代码：

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <cmath>using namespace std;struct Point{ double x,y;};struct Segment{ Point p1,p2;};double cross(const Point& o,const Point & a,const Point& b){ return (a.x - o.x)*(b.y-o.y) - (a.y - o.y)* (b.x - o.x);}double dot(const Point& o,const Point & a,const Point& b){ return (a.x - o.x) * (b.x - o.x) + (a.y-o.y) * (b.y - o.y);}const double eps = 1E-8;int sig(double d){ return (d > eps) - (d < -eps);}int btw(const Point& x,const Point& a,const Point&b){ return sig(dot(x,a,b));}int SegCross(const Point & a,const Point & b,const Point& c,const Point& d,Point &p){ int d1,d2,d3,d4; double s1,s2; d1 = sig(s1 = cross(a,b,c)); d2 = sig(s2 = cross(a,b,d)); d3 = sig(cross(c,d,a)); d4 = sig(cross(c,d,b)); if((d1^d2)==-2&&(d3^d4)==-2) { p.x = (c.x * s2 - d.x * s1)/(s2 - s1); p.y = (c.y * s2 - d.y * s1)/(s2 - s1); return 1; } if(d1 == 0 && d2 != 0&& btw(c,a,b)<=0) { p = c; return 2; } if(d2 == 0 && d1 != 0&& btw(d,a,b)<=0 ) { p = d; return 2; } if(d3 == 0 && d4 != 0&& btw(a,c,d)<=0 ) { p = a; return 2; } if(d4 == 0 && d3 != 0&& btw(b,c,d) <=0) { p = b; return 2; } return 0;}bool Find(const Point& a,const Point& b,const Point&c ,const Point& d ,const Point& p, Point& p1, Point&p2){ bool find = false; if(a.y > p.y ) { find = true; p1 = a; } if(b.y > p.y) { if(find ) { if(b.y > p1.y) p1 = b; } else p1 = b; find = true; } if(c.y > p.y) { if(find ) { if(c.y > p1.y) p1 = c; } else p1 = c; find = true; } if(d.y > p.y) { if(find ) { if(d.y > p1.y) p1 = d; } else p1 = d; find = true; } //cout << "find = " << find << " " << p1.x << " " << p1.y << endl;; if(find) { find = false; if(a.y > p.y && (sig(a.x - p1.x)!= 0 || sig(a.y - p1.y)!= 0)) { find = true; p2 = a; } if(b.y > p.y && (sig(b.x - p1.x)!= 0 || sig(b.y - p1.y)!= 0)) { if(find ) { if(b.y > p2.y) p2 = b; } else p2 = b; find = true; } if(c.y > p.y && (sig(c.x - p1.x)!= 0 || sig(c.y - p1.y)!= 0)) { if(find ) { if(c.y > p2.y) p2 = c; } else p2 = c; find = true; } if(d.y > p.y && (sig(d.x - p1.x)!= 0 || sig(d.y - p1.y)!= 0)) { if(find ) { if(d.y > p2.y) p2 = d; } else p2 = d; find = true; } } return find;}int SegLineCross(const Point & a,const Point & b,const Point & c,const Point & d,Point & p){ double s1,s2; int d1,d2; d1 = sig(s1 = cross(a,b,c)); d2 = sig(s2 = cross(a,b,d)); //cout << "d1 = " << d1 << endl; //cout << "d2 = " << d2 << endl; if((d1^d2)==-2) { p.x = (c.x * s2 - d.x * s1)/(s2 - s1); p.y = (c.y * s2 - d.y * s1)/(s2 - s1); return 1; } if(d1 == 0 && d2 != 0) { p = c; return 2; } if(d1 != 0 && d2 == 0) { p = d; return 2; } return 0;}int main(){ int n; cin >> n; Segment seg1,seg2; Point p,p1,p2,p3; Point tp; while(n--) { scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&seg1.p1.x,&seg1.p1.y,&seg1.p2.x,&seg1.p2.y, &seg2.p1.x,&seg2.p1.y,&seg2.p2.x,&seg2.p2.y); if(SegCross(seg1.p1,seg1.p2,seg2.p1,seg2.p2,p)==0) { cout << "0.00" << endl; } else { Find(seg1.p1,seg1.p2,seg2.p1,seg2.p2,p,p1,p2);// cout <<" p = " << p.x << " " << p.y << endl;// cout <<" tp = " << tp.x << " " << tp.y << endl;// cout <<" p1 = " << p1.x << " " << p1.y << endl;// cout <<" p2 = " << p2.x << " " << p2.y << endl;// cout <<" p3 = " << p3.x << " " << p3.y << endl; if(Find(seg1.p1,seg1.p2,seg2.p1,seg2.p2,p,p1,p2)) { tp.x = p2.x; tp.y = p2.y + 1; // cout << "SegLineCross(p2,tp,p,p1,p3) == " << SegLineCross(p2,tp,p,p1,p3) << endl; if(SegLineCross(p2,tp,p,p1,p3)!=0) { if(p3.y > p2.y) cout << "0.00" << endl; else { tp.y = p2.y; tp.x = p2.x + 1; SegLineCross(p2,tp,p,p1,p3); printf("%.2f/n",fabs(cross(p,p3,p2))/2); } } else { tp.y = p2.y; tp.x = p2.x + 1; SegLineCross(p2,tp,p,p1,p3);// cout <<" p = " << p.x << " " << p.y << endl;// cout <<" tp = " << tp.x << " " << tp.y << endl;// cout <<" p1 = " << p1.x << " " << p1.y << endl;// cout <<" p2 = " << p2.x << " " << p2.y << endl;// cout <<" p3 = " << p3.x << " " << p3.y << endl; printf("%.2f/n",fabs(cross(p,p3,p2))/2); } //printf("%.2f/n",fabs(cross(p,p1,p2))/2); } else cout << "0.00" << endl; } } return 0;}