HDU4404 Worms

http://acm.hdu.edu.cn/showproblem.php?pid=4404

2012jinhua online

求圆与多边形相交的面积。

模板题。

#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <set>//#include <vector>#include <iostream>#include <algorithm>#include<string>using namespace std;//const double eps=1e-7;//const double INF=1e50;//const double pi=acos(-1);const int maxn = 105;#define pi  acos(-1.0)struct point1{    double x, y ;    point1(){}    point1(double x, double y):x(x),y(y){}    double operator *(const point1 &b)const{        return x * b.y - y * b.x;    }    point1 operator -(const point1 &b)const{        return point1( x - b.x, y - b.y );    }    void in(){         scanf("%lf%lf",&x,&y);    }}p[maxn] , q[maxn], o;const double eps = 1e-10;inline double max (double a, double b) { if (a > b) return a; else return b; }inline double min (double a, double b) { if (a < b) return a; else return b; }inline int fi (double a){    if (a > eps) return 1;    else if (a >= -eps) return 0;    else return -1;}class vector{public:    double x, y;    vector (void) {}    vector (double x0, double y0) : x(x0), y(y0) {}    double operator * (const vector& a) const { return x * a.y - y * a.x; }    double operator % (const vector& a) const { return x * a.x + y * a.y; }    vector verti (void) const { return vector(-y, x); }    double length (void) const { return sqrt(x * x + y * y); }    vector adjust (double len)    {        double ol = len / length();        return vector(x * ol, y * ol);    }    vector oppose (void) { return vector(-x, -y); }};class point{public:    double x, y;    point (void) {}    point (double x0, double y0) : x(x0), y(y0) {}    vector operator - (const point& a) const { return vector(x - a.x, y - a.y); }    point operator + (const vector& a) const { return point(x + a.x, y + a.y); }};class segment{public:    point a, b;    segment (void) {}    segment (point a0, point b0) : a(a0), b(b0) {}    point intersect (const segment& s) const    {        vector v1 = s.a - a, v2 = s.b - a, v3 = s.b - b, v4 = s.a - b;        double s1 = v1 * v2, s2 = v3 * v4;        double se = s1 + s2;        s1 /= se, s2 /= se;        return point(a.x * s2 + b.x * s1, a.y * s2 + b.y * s1);    }    point pverti (const point& p) const    {        vector t = (b - a).verti();        segment uv(p, p + t);        return intersect(uv);    }    bool on_segment (const point& p) const    {        if (fi(min(a.x, b.x) - p.x) <= 0 && fi(p.x - max(a.x, b.x)) <= 0 &&            fi(min(a.y, b.y) - p.y) <= 0 && fi(p.y - max(a.y, b.y)) <= 0) return true;        else return false;    }};double radius;point polygon[550];double kuras_area (point a, point b){    point ori(0, 0);    int sgn = fi((b - ori) * (a - ori));    double da = (a - ori).length(), db = (b - ori).length();    int ra = fi(da - radius), rb = fi(db - radius);    double angle = acos(((b - ori) % (a - ori)) / (da * db));    segment t(a, b); point h, u; vector seg;    double ans, dlt, mov, tangle;    if (fi(da) == 0 || fi(db) == 0) return 0;    else if (sgn == 0) return 0;    else if (ra <= 0 && rb <= 0) return fabs((b - ori) * (a - ori)) / 2 * sgn;    else if (ra >= 0 && rb >= 0)    {        h = t.pverti(ori);        dlt = (h - ori).length();        if (!t.on_segment(h) || fi(dlt - radius) >= 0)            return radius * radius * (angle / 2) * sgn;        else        {            ans = radius * radius * (angle / 2);            tangle = acos(dlt / radius);            ans -= radius * radius * tangle;            ans += radius * sin(tangle) * dlt;            return ans * sgn;        }    }    else    {        h = t.pverti(ori);        dlt = (h - ori).length();        seg = b - a;        mov = sqrt(radius * radius - dlt * dlt);        seg = seg.adjust(mov);        if (t.on_segment(h + seg)) u = h + seg;        else u = h + seg.oppose();        if (ra == 1) swap(a, b);        ans = fabs((a - ori) * (u - ori)) / 2;        tangle = acos(((u - ori) % (b - ori)) / ((u - ori).length() * (b - ori).length()));        ans += radius * radius * (tangle / 2);        return ans * sgn;    }}int main(){    //freopen("a","r",stdin);    int n, i;    double v0,thi,time1,g,r;    while(1)    {        scanf("%lf%lf%lf%lf%lf%lf%lf",&o.x,&o.y,&v0,&thi,&time1,&g,&radius);        if (fabs(o.x)<eps && fabs(o.y)<eps && fabs(v0)<eps &&  fabs(thi)<eps && fabs(time1)<eps && fabs(g)<eps && fabs(radius)<eps) break;        thi=thi*acos(-1)/180.00;        o.x += v0*cos(thi)*time1 , o.y += v0*sin(thi)*time1-0.5*g*time1*time1 ;       // cout<<o.x<<' '<<o.y<<' '<<r<<endl;        double x,y;        scanf("%d",&n);        for (i = 0; i < n; i++)        {            scanf("%lf %lf", &x, &y);            x-=o.x;            y-=o.y;            polygon[i] = point(x, y);        }        double area = 0;        for (int i = 0; i < n; i++)            area += kuras_area(polygon[i], polygon[(i + 1) % n]);        printf("%.2f\n", fabs(area));    }    return 0;}

#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <set>//#include <vector>#include <iostream>#include <algorithm>#include<string>using namespace std;//const double eps=1e-7;//const double INF=1e50;//const double pi=acos(-1);const int maxn = 105;#define pi  acos(-1.0)#define eps 1e-7// poj 3675 一个圆和一个多边形的面积交的大double sign( double x ){    if( x < -eps ) return -1;    return x > eps;}struct point{    double x, y ;    point(){}    point(double x, double y):x(x),y(y){}    double operator *(const point &b)const{        return x * b.y - y * b.x;    }    point operator -(const point &b)const{        return point( x - b.x, y - b.y );    }    void in(){         scanf("%lf%lf",&x,&y);    }}p[maxn] , q[maxn], o,p0;double r ,ans ; // 已原点为圆心的圆半径double thi,time1,g,v0;////圆o与线段x-y的交点,,如果存在顺序存入p[0]和p[1],,返回交点的个数int xianduan_jiao_yuan( point o, double r,point u, point v, point &o1, point &o2){    u = u - o;  v = v - o;    point  w = v - u ;    double a = w.x * w.x + w.y * w.y ;    double b = w.x * u.x * 2 + w.y * u.y * 2;    double c = u.x * u.x + u.y * u.y - r * r;    double d = b * b - 4 * a * c;    if( sign( d ) < 0) return 0;    d = sqrt( d );    double t1 = ( - b + d )/2/a, t2 = ( - b - d )/2/a;    if( t1 > t2 )  swap( t1, t2 );    int ans = 0;    ////注意这里端点在圆上没有算作相交,有需要改成等号既可    if( sign( t1 ) > 0 && sign( t1 - 1 ) < 0 ){        ans ++ ;        o1 = point( u.x + w.x * t1 + o.x, u.y + w.y*t1+o.y);    }////注意重根这里算作了一个交点,,    if(sign( t1 - t2) != 0 && sign(t2)>0&&sign(t2-1)<0){        ans ++;        if( ans == 1 ) o1 = point( u.x+w.x*t2+o.x,u.y+w.y*t2+o.y );        else o2 = point( u.x+w.x*t2+o.x,u.y+w.y*t2+o.y );    }    return ans ;}double get( point p[], int n ){     double ans = 0;     for( int i = 1; i < n; i ++ ){          point a, b;          a = p[ i - 1]; b = p[ i ];//只要有一个人在圆外面 那就只需要计算扇形的面积即可          if( sqrt(a.x*a.x+a.y*a.y) - r > eps || sqrt( b.x * b.x + b.y*b.y) - r > eps ){              double t = atan2(b.y , b.x ) - atan2(a.y , a.x ) ;              while( t - pi > eps  ) t -= pi * 2 ;//t的范围是( -pi , pi )              while( t + pi < -eps ) t += pi * 2 ;              ans += t * r * r ;          }else ans += a * b ;  //叉乘  这个在圆里面     }     return ans ;}void add( point a , point b ){     point p1, p2;     int end = 0;     p[end++] = a;     point o1, o2;     int t = xianduan_jiao_yuan(point(0,0),r,a,b,o1,o2);     if( t >= 1) p[end++] = o1;     if( t >= 2) p[end++] = o2;     p[end++] = b;     ans += get( p, end );}int main(){    //freopen("a","r",stdin);    int n, i;    double v0,thi,time1,g;    while(1)    {        scanf("%lf%lf%lf%lf%lf%lf%lf",&o.x,&o.y,&v0,&thi,&time1,&g,&r);        if (fabs(o.x)<eps && fabs(o.y)<eps && fabs(v0)<eps &&  fabs(thi)<eps && fabs(time1)<eps && fabs(g)<eps && fabs(r)<eps) break;        thi=thi*acos(-1)/180.00;        o.x += v0*cos(thi)*time1 , o.y += v0*sin(thi)*time1-0.5*g*time1*time1 ;        ans=0;        scanf("%d",&n);        for(  i = 0; i < n; i ++ ){            scanf("%lf%lf",&q[i].x, &q[i].y );            q[i] = q[i] - o;        }        q[n] = q[0];        for( i  =0; i < n; i++ ){            add( q[i], q[i+1] );        }        printf("%.2lf\n",fabs(ans)/2);    }    return 0;}