1845: [Cqoi2005] 三角形面积并
Time Limit: 3 Sec Memory Limit: 64 MB
Submit: 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
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;
}
时间： 2024-12-28 20:05:30