bzoj 1845: [Cqoi2005] 三角形面积并 扫描线
1845: [Cqoi2005] 三角形面积并
Time Limit: 3 Sec Memory Limit: 64 MBSubmit: 848 Solved: 206
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Description
给出n个三角形,求它们并的面积。
Input
第一行为n(N < = 100), 即三角形的个数 以下n行,每行6个整数x1, y1, x2, y2, x3, y3,代表三角形的顶点坐标。坐标均为不超过10 ^ 6的实数,输入数据保留1位小数
Output
输出并的面积u, 保留两位小数
Sample Input
2
0.0 0.0 2.0 0.0 1.0 1.0
1.0 0.0 3.0 0.0 2.0 1.0
0.0 0.0 2.0 0.0 1.0 1.0
1.0 0.0 3.0 0.0 2.0 1.0
Sample Output
1.75
以前一直听说有一个除辛普森积分外的求面积的方法,现在才终于编了一次,大概思路就是通过扫描线,将答案转换成一个一个梯形相加。
另:unique的函数参数应该传进去等于符号的比较函数,而不是大于符号,每次都会搞错。。
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #include<cmath> using namespace std; #define MAXN 1000 #define sqr(x) ((x)*(x)) #define eps 1e-10 typedef double real; inline int sgn(real x) { if (abs(x)<eps) return 0; return x<0?-1:1; } struct point { real x,y; point(real x,real y):x(x),y(y){} point(){} void read() { scanf("%lf%lf",&x,&y); } }; real dis(point p1,point p2) { return sqrt(sqr(p1.x-p2.x)+sqr(p1.y-p2.y)); } struct line { point ps; real x,y; line(){} line(point p1,point p2) { ps=p1; x=p2.x-p1.x; y=p2.y-p1.y; } point spos() { return ps; } point tpos() { return point(ps.x+x,ps.y+y); } bool inside(point pt) { return sgn(sqrt(x*x+y*y)-dis(ps,pt)-dis(tpos(),pt))==0; } point get_point(real xx) { return point(xx,ps.y+y/x*(xx-ps.x)); } void operator *=(real k) { x*=k;y*=k; } void print() { printf("Line:[%.2lf,%.2lf]->[%.2lf,%.2lf]\n",ps.x,ps.y,ps.x+x,ps.y+y); } }lst[MAXN]; int topl=-1; real xmul(line l1,line l2) { return l1.x*l2.y-l1.y*l2.x; } line operator -(line l1) { l1.ps=l1.tpos(); l1.x=-l1.x; l1.y=-l1.y; return l1; } bool parallel(line l1,line l2) { return !sgn(xmul(l1,l2)); } point crossover(line l1,line l2) { real s1=-xmul(line(l2.spos(),l1.spos()),l1); real s2=xmul(line(l2.tpos(),l1.spos()),l1); return point(l2.ps.x+l2.x*s1/(s1+s2), l2.ps.y+l2.y*s1/(s1+s2)); } point pl[MAXN*MAXN]; int topp=-1; bool cmp_x(point p1,point p2) { return sgn(p1.x-p2.x)<0; } bool equal_x(point p1,point p2) { return sgn(p1.x-p2.x)==0; } line seq[MAXN]; bool cmp_line(line l1,line l2) { point p1,p2; if (l1.x>=0) p1=l1.spos(); else p1=l1.tpos(); if (l2.x>=0) p2=l2.spos(); else p2=l2.tpos(); if (sgn(p1.y-p2.y)==0) { if (l1.x>=0) p1=l1.tpos(); else p1=l1.spos(); if (l2.x>=0) p2=l2.tpos(); else p2=l2.spos(); return p1.y>p2.y; }else { return p1.y>p2.y; } } int main() { freopen("input.txt","r",stdin); int n; point p1,p2,p3; scanf("%d",&n); for (int i=0;i<n;i++) { p1.read(); p2.read(); p3.read(); lst[++topl]=line(p1,p2); lst[++topl]=line(p2,p3); lst[++topl]=line(p3,p1); if (xmul(-lst[topl-1],lst[topl])<0) { lst[topl]=-lst[topl]; lst[topl-1]=-lst[topl-1]; lst[topl-2]=-lst[topl-2]; } } point pt; for (int i=0;i<=topl;i++) { for (int j=i+1;j<=topl;j++) { pt=crossover(lst[i],lst[j]); if (lst[i].inside(pt) && lst[j].inside(pt)) pl[++topp]=pt; } } sort(pl,pl+topp+1,cmp_x); topp=unique(pl,pl+topp+1,equal_x)-pl-1;//*** real a,b; line lt; real ans=0; for (int i=1;i<=topp;i++) { a=pl[i-1].x; b=pl[i].x; int tops=-1; for (int j=0;j<=topl;j++) { lt=lst[j]; if ((lt.spos().x<lt.tpos().x && sgn(lt.spos().x-a)<=0 && sgn(b-lt.tpos().x)<=0) || (lt.spos().x>lt.tpos().x && sgn(lt.tpos().x-a)<=0 && sgn(b-lt.spos().x)<=0)) { if (lt.spos().x<lt.tpos().x) { lt.ps=lt.get_point(a); lt*=(b-a)/lt.x; seq[++tops]=lt; }else { lt.ps=lt.get_point(b); lt*=(a-b)/lt.x; seq[++tops]=lt; } } } sort(seq,seq+tops+1,cmp_line); /* printf("Segment:%.2lf %.2lf\n",a,b); for (int j=0;j<=tops;j++) seq[j].print();*/ int cnt=0; for (int j=0;j<=tops;j++) { if (seq[j].x>=0) { cnt++; if (cnt==1) ans+=(seq[j].spos().y+seq[j].tpos().y)*seq[j].x/2; } else { cnt--; if (cnt==0) ans+=(seq[j].spos().y+seq[j].tpos().y)*seq[j].x/2; } } } printf("%.2lf\n",ans); return 0; }
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