HDU 2892 area （多边形和圆面积并--基础题）

【题目链接】：click here~~

【题目大意】：

Description

小白最近被空军特招为飞行员，参与一项实战演习。演习的内容是轰炸某个岛屿。。。 
作为一名优秀的飞行员，任务是必须要完成的，当然，凭借小白出色的操作，顺利地将炸弹投到了岛上某个位置，可是长官更关心的是，小白投掷的炸弹到底摧毁了岛上多大的区域？ 
岛是一个不规则的多边形，而炸弹的爆炸半径为R。 
小白只知道自己在（x，y，h）的空间坐标处以（x1,y1,0）的速度水平飞行时投下的炸弹，请你计算出小白所摧毁的岛屿的面积有多大. 重力加速度G = 10. 

 

Input

首先输入三个数代表小白投弹的坐标（x，y，h）； 
然后输入两个数代表飞机当前的速度（x1, y1）; 
接着输入炸弹的爆炸半径R； 
再输入一个数n，代表岛屿由n个点组成； 
最后输入n行，每行输入一个（x'，y'）坐标，代表岛屿的顶点（按顺势针或者逆时针给出）。(3<= n < 100000) 

 

Output

输出一个两位小数，表示实际轰炸到的岛屿的面积。

 

Sample Input

     
     
     
     

      
      
      
      0 0 2000 100 0 100 4 1900 100 2000 100 2000 -100 1900 -100 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
      15707.96 
     
     
     
     

 

Source

2009 Multi-University Training Contest 10 - Host by NIT

【思路】：如果忘了抛物线的一些公式，建议上百度复习一下，这里我们假设最后投掷点圆心坐标为(ceter_x0,ceter_y0)，空间坐标（x,y,H）,初始速度（x1,y1,0）那么由高度为H，最后投掷点到地面竖直速度为0，由公式得：起点速度为：V^2=2*g*H,得 V=sqrt(2*g*H),那么时间t为V/g=sqrt(2*g*H)/g,注意这里是空间坐标，那么最后ceter_x0=x+t*x1,ceter_y0=y+t*y1。

由此，套入模板求出面积。

代码：

/*
* 多边形和圆面积并
* Problem: HDU No.2892
* Running time: 62MS
* Complier: G++
* Author: javaherongwei
* Create Time: 15:34 2015/10/1 星期四
*/
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi=acos(-1.0);
const double e=exp(1.0);
const double eps=1e-8;
const int maxn=100005;
double R,k;
int n,m;
struct point          // 点或向量结构
{
    double x,y;
    point(double _x=0.0,double _y=0.0):x(_x),y(_y) {}
    point operator - (const point &p)
    {
        return point(x-p.x,y-p.y);
    }
    double sqrx()    //向量的模
    {
        return sqrt(x*x+y*y);
    }
} area[maxn];

int dcmp(double x)
{
    return (x>eps)-(x<-eps);
}

double xmult(point &p1,point &p2,point &p0)//叉积
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}

double distancex(point &p1,point &p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}

point intersection(point u1,point u2,point v1,point v2)   //两直线交点
{
    point ret = u1;
    double t = ((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
    ret.x += (u2.x-u1.x)*t;
    ret.y += (u2.y-u1.y)*t;
    return ret;
}

void intersection_line_circle(point c, double r, point l1, point l2, point & p1, point & p2) //直线与圆相交
{
    point p = c;
    double t;
    p.x += l1.y-l2.y;
    p.y += l2.x-l1.x;
    p = intersection(p, c, l1, l2);
    t = sqrt(r*r-distancex(p, c)*distancex(p, c))/distancex(l1, l2);
    p1.x = p.x+(l2.x-l1.x)*t;
    p1.y = p.y+(l2.y-l1.y)*t;
    p2.x = p.x-(l2.x-l1.x)*t;
    p2.y = p.y-(l2.y-l1.y)*t;
}

point len_pot_seg(point p, point l1, point l2)//点到线段的最近距离
{
    point t = p;
    t.x += l1.y-l2.y;
    t.y += l2.x-l1.x;
    if (xmult(l1, t, p)*xmult(l2, t, p)>eps)
        return distancex(p, l1)<distancex(p, l2) ? l1 : l2;
    return intersection(p, t, l1, l2);
}

double distp(point & a, point & b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}

double Direct_Triangle_Circle_Area(point a, point b, point o, double r)
{
    double sign = 1.0;
    a = a-o;
    b = b-o;
    o = point(0.0, 0.0);
    if(fabs(xmult(a, b, o)) < eps) return 0.0;
    if(distp(a, o) > distp(b, o))
    {
        swap(a, b);
        sign = -1.0;
    }
    if (distp(a, o) < r*r+eps)
    {
        if (distp(b, o) < r*r+eps) return xmult(a, b, o)/2.0*sign;
        point p1, p2;
        intersection_line_circle(o, r, a, b, p1, p2);
        if (distancex(p1, b) > distancex(p2, b)) swap(p1, p2);
        double ret1 = fabs(xmult(a, p1, o));
        double ret2 = acos((p1.x*b.x+p1.y*b.y)/p1.sqrx()/b.sqrx())*r*r;
        double ret = (ret1+ret2)/2.0;
        if (xmult(a, b, o)<eps && sign>0.0 || xmult(a, b, o)>eps && sign<0.0) ret = -ret;
        return ret;
    }
    point ins = len_pot_seg(o, a, b);
    if(distp(o, ins)>r*r-eps)
    {
        double ret = acos((a.x*b.x+a.y*b.y)/a.sqrx()/b.sqrx())*r*r/2.0;
        if(xmult(a, b, o)<eps && sign>0.0 || xmult(a, b, o)>eps && sign<0.0) ret = -ret;
        return ret;
    }
    point p1, p2;
    intersection_line_circle(o, r, a, b, p1, p2);
    double cm = r/(distancex(o, a)-r);
    point m = point((o.x+cm*a.x)/(1+cm),(o.y+cm*a.y)/(1+cm));
    double cn = r/(distancex(o, b)-r);
    point n = point((o.x+cn*b.x)/(1+cn),(o.y+cn*b.y)/(1+cn));
    double ret1 = acos((m.x*n.x+m.y*n.y)/m.sqrx()/n.sqrx())*r*r;
    double ret2 = acos((p1.x*p2.x+p1.y*p2.y)/p1.sqrx()/p2.sqrx())*r*r-fabs(xmult(p1, p2, o));
    double ret = (ret1-ret2)/2.0;
    if(xmult(a, b, o)<eps && sign>0.0||xmult(a, b, o)>eps && sign<0.0) ret=-ret;
    return ret;
}

int main()
{
    int tot=1;
    double x,y,H;
    double x1,y1,R;
    while(cin>>x>>y>>H)
    {
        cin>>x1>>y1;
        cin>>R;
        cin>>n;
        for(int i=0; i<n; i++) scanf("%lf%lf", &area[i].x, &area[i].y);
        double V=sqrt(2*10*H);
        double t=V/10;
        double ceter_x0=x+x1*t;
        double ceter_y0=y+y1*t;
        point ceter_xy=point(ceter_x0,ceter_y0);
        double sum=0.0;
        for(int i=0; i<n-1; i++) sum += Direct_Triangle_Circle_Area(area[i], area[i+1], ceter_xy, R);
        sum += Direct_Triangle_Circle_Area(area[n-1], area[0], ceter_xy, R);
        printf("%.2f\n",fabs(sum));
    } return 0;
}