poj2986A Triangle and a Circle&&poj3675Telescope（三角形剖分）

链接

2986是3675的简化版，只有一个三角形。

这题主要在于求剖分后三角形与圆的相交面积，需要分情况讨论。

具体可以看此博客 http://hi.baidu.com/billdu/item/703ad4e15d819db52f140b0b

在分析第3、4两种情况时，我是用角度来进行判断的，如果<obc||<ocb大于90度就为他所说的第四种情况，不然就是第三种情况。

还有对于sig的解释貌似网上都没写，可能都觉得太简单了。。。自己手画了一下，大体是这个样子的

红色标记那块三角形是需要减掉对于当前多边形，可以看出以最下角进行剖分三角形时，cross(b,c)算的那块小三角形的确是负的，所以需要判断一下当前的面积是要加上的还是要减掉的。

讨论的东西比较多，细节比较多，WA了好多遍，对着数据查了好久终于过了。。

附上一些数据

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 100
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y) {}
} p[N];
struct tri
{
    point a,b,c;
} tr[N];
typedef point pointt;
point operator -(point a,point b)
{
    return point(a.x-b.x,a.y-b.y);
}
point operator *(point a,double r)
{
    return point(a.x*r,a.y*r);
}
point operator +(point a,point b)
{
    return point(a.x+b.x,a.y+b.y);
}
struct line
{
    point u,v;
    point ppoint(double t)
    {
        return point(u+v*t);
    }
};
struct circle
{
    point c;
    double r;
    circle(point c,double r):c(c),r(r) {}
    point ppoint(double a)
    {
        return point(c.x+cos(a)*r,c.y+sin(a)*r);
    }
};
double r;
point ip;
double dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}
double dis(point a)
{
    return sqrt(a.x*a.x+a.y*a.y);
}
double dot(point a,point b)
{
    return a.x*b.x+a.y*b.y;
}
double cross(point a,point b)
{
    return a.x*b.y-a.y*b.x;
}
double area(point a,point b,point c)
{
    return fabs(cross(a-c,b-c))/2;
}

int getlinecircle(line ll,circle cc,point &p1,point &p2)
{
    double a = ll.v.x,b = ll.u.x-cc.c.x,c = ll.v.y,d = ll.u.y-cc.c.y;
    double e = a*a+c*c,f = 2*(a*b+c*d),g = b*b+d*d-cc.r*cc.r;
    double delta = f*f-4*e*g;
    double t1,t2;
    if(dcmp(delta)<0)return 0;//ÏàÀë
    if(dcmp(delta)==0)
    {
        t1 = t2 = -f/(2*e);//cout<<t1<<" -"<<e<<" "<<f<<endl;
        p1 = ll.ppoint(t1);
        return 1;//ÏàÇÐ
    }
    //Ïཻ
    t1 = (-f-sqrt(delta))/(2*e);
    p1 = ll.ppoint(t1);
    t2 = (-f+sqrt(delta))/(2*e);
    p2 = ll.ppoint(t2);
    // cout<<p1.x<<" "<<p1.y<<" "<<p2.x<<" "<<p2.y<<endl;
    return 2;
}
double mul(point a,point b,point c)
{
    return cross(b-a,c-a);
}
bool cmp(point a,point b)
{
    if(dcmp(mul(ip,a,b))==0)
        return dis(a-ip)<dis(b-ip);
    else
        return dcmp(mul(ip,a,b))>0;
}
double distancetoline(point p,point a,point b)
{
    point v1 = a-b,v2 = p-b;
    return fabs(cross(v1,v2))/dis(v1);
}
int dot_online_in(point p,point l1,point l2)
{
    return !dcmp(mul(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
}
double angle(point a,point b)
{
    return acos(dot(a,b)/dis(a)/dis(b));
}
double cal(tri tr)
{
    circle cp=circle(point(0,0),r);
    int sig = dcmp(cross(tr.b,tr.c));
    if(sig==0) return 0;
    double d1 = dis(tr.a-tr.b),d2 = dis(tr.a-tr.c);
    if(dcmp(d1-r)<=0&&dcmp(d2-r)<=0)
    {
        double s = sig*area(tr.a,tr.b,tr.c);
        return s;
    }
    double dline = distancetoline(cp.c,tr.b,tr.c);
    if(dcmp(d1-r)>=0&&dcmp(d2-r)>=0&&dcmp(dline-r)>=0)
    {
        return sig*angle(tr.b,tr.c)*r*r/2.0;
    }
    double ag = angle(tr.c-tr.b,tr.a-tr.b),bg = angle(tr.b-tr.c,tr.a-tr.c);
    point p1,p2;
    line l1;
    l1.u = tr.b,l1.v = tr.c-tr.b;
    getlinecircle(l1,cp,p1,p2);

    if(dcmp(d1-r)>=0&&dcmp(d2-r)>=0&&dcmp(dline-r)<0&&(dcmp(ag-pi/2)>=0||dcmp(bg-pi/2)>=0))
    {

        double s = sig*angle(tr.b,tr.c)*r*r/2;
        return s;
    }
    if(dcmp(d1-r)>=0&&dcmp(d2-r)>=0&&dcmp(dline-r)<0)
    {
        double s = (angle(tr.b,tr.c)-angle(p1,p2))*r*r/2.0+area(tr.a,p1,p2);
        return sig*s;
    }

    p1 = dot_online_in(p1,tr.b,tr.c)?p1:p2;
    if(dcmp(d1-r)<0)
    {
        return sig*(angle(tr.c,p1)*r*r/2+area(tr.a,p1,tr.b));
    }
    else
    {
        return sig*(angle(p1,tr.b)*r*r/2+area(tr.a,p1,tr.c));
    }
}
int dots_inline(point p1,point p2,point p3)
{
    return !dcmp(mul(p1,p2,p3));
}
int main()
{
    int i,n;
    while(scanf("%lf",&r)!=EOF)
    {
        scanf("%d",&n);
        for(i = 0; i < n ; i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        p[n] = p[0];
        double ans = 0;
        for(i = 0 ; i < n ; i++)
        {
            if(dots_inline(ip,p[i],p[i+1])) continue;
            tr[i].a = point(0,0);
            tr[i].b = p[i];
            tr[i].c = p[i+1];
            ans+=cal(tr[i]);
        }
        printf("%.2f
",fabs(ans)+eps);
    }
    return 0;
}

589.00 191.00 -554.00 710.00 748.00 774.00 -888.00 -588.00 902.00

201.00 -847.00 -365.00 886.00 -557.00 -609.00 272.00 -345.00 189.00

-358.00 981.00 269.00 511.00 158.00 -304.00 468.00 463.00 834.00

969.00 514.00 -445.00 460.00 -177.00 774.00 -34.00 -125.00 162.00

-467.00 413.00 -714.00 -986.00 362.00 666.00 813.00 271.00 264.00

-497.00 908.00 -414.00 631.00 -220.00 868.00 166.00 -258.00 306.00

-107.00 -743.00 -952.00 322.00 -273.00 -214.00 -14.00 466.00 758.00

511.00 -416.00 -934.00 -745.00 -335.00 -132.00 -482.00 391.00 626.00

928.00 821.00 -293.00 -853.00 -488.00 -312.00 -27.00 94.00 361.00

-979.00 -280.00 791.00 -943.00 -300.00 -278.00 -821.00 684.00 365.00

-700.00 955.00 -315.00 154.00 -103.00 -606.00 404.00 -792.00 940.00

607.00 783.00 597.00 944.00 -672.00 -323.00 343.00 -799.00 526.00

815.00 -390.00 -291.00 37.00 422.00 687.00 672.00 613.00 848.00

-988.00 363.00 -529.00 660.00 -597.00 143.00 502.00 459.00 522.00

-206.00 484.00 109.00 -111.00 424.00 650.00 330.00 -545.00 480.00

94.00 -638.00 -59.00 -9.00 -400.00 -702.00 0.00 267.00 741.00

-859.00 522.00 109.00 -640.00 383.00 712.00 489.00 -663.00 635.00

808.00 -31.00 471.00 172.00 -374.00 21.00 120.00 -860.00 474.00

-539.00 -887.00 498.00 844.00 -453.00 -213.00 -479.00 -9.00 315.00

答案

Case 1
0.00
Case 2
0.00
Case 3
274955.27
Case 4
0.00
Case 5
0.00
Case 6
0.00
Case 7
25157.17
Case 8
9943.87
Case 9
181113.99
Case 10
0.00
Case 11
11846.16
Case 12
0.00
Case 13
404668.37
Case 14
0.00
Case 15
0.00
Case 16
74663.53
Case 17
80015.79
Case 18
0.00
Case 19
57316.85
Case 20
0.00