POJ 2451 Uyuw's Concert 题解

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#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

const int N = 20009;
const double eps = 1e-8;
const double maxl = 70000;
const double pi = acos(-1.0);

struct cpoint {//C++构造函数,默认缺省值为(0,0)
double x, y;
cpoint(double xx = 0, double yy = 0): x(xx), y(yy) {};
};

int dcmp(double x) {//判断参数的符号,负数返回-1,0返回0,正数返回1
if (x < -eps) return -1;
else return x > eps;
}

double xmult(cpoint p0, cpoint p1, cpoint p2) { // p0p1 与 p0p2 叉积
return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}

bool EqualPoint(cpoint a, cpoint b) {//两点相等
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

struct cvector {//向量
cpoint s, e;
double ang, d;
};

//atan (double x)弧度表示的反正切
//atan2(double x,double y)弧度表示的反正切,相当于atan(x/y)

void setline(double x1, double y1, double x2, double y2, cvector &v) {
v.s.x = x1; v.s.y = y1;
v.e.x = x2; v.e.y = y2;
v.ang = atan2(y2 - y1, x2 - x1);
//这里的 d 代表向量(直线)和坐标轴的截距,正数表示 原点 在该向量的左边
//(这道题要求左半平面交),负号则表示 原点 在右边
if (dcmp(x1 - x2))          // x1 > x2
v.d = (x1 * y2 - x2 * y1) / fabs(x1 - x2);
else v.d = (x1 * y2 - x2 * y1) / fabs(y1 - y2);
}

//判向量平行
bool parallel(const cvector &a, const cvector &b) {
double u = (a.e.x - a.s.x) * (b.e.y - b.s.y) - (a.e.y - a.s.y) * (b.e.x - b.s.x);
return dcmp(u) == 0;
}

//求两向量(直线)交点 (两向量不能平行或重合)
cpoint CrossPoint(const cvector &a, const cvector &b) {
cpoint res;
double u = xmult(a.s, a.e, b.s), v = xmult(a.e, a.s, b.e);
res.x = (b.s.x * v + b.e.x * u) / (u + v);
res.y = (b.s.y * v + b.e.y * u) / (u + v);
return res;
}

//半平面交排序函数 [优先顺序: 1.极角 2.前面的直线在后面的左边]
static bool VecCmp(const cvector &l, const cvector &r) {
if (dcmp(l.ang - r.ang)) return l.ang < r.ang;
return l.d < r.d;
}

cvector deq
; //用于计算的双端队列

void HalfPanelCross(cvector vec[],int n,cpoint cp[],int &m)
{
int i,tn;
sort(vec,vec+n,VecCmp);
for(tn=i=1;i<n;i++)//去掉极角相同的边，同时留下最右边的
{
if(dcmp(vec[i].ang-vec[i-1].ang) != 0)
vec[tn++]=vec[i];
}n=tn;
int bot=0, top=1;
deq[0]=vec[0], deq[1]=vec[1];
for(i=2;i<tn;i++)
{
if(parallel(deq[top],deq[top-1]) || parallel(deq[bot],deq[bot+1]))return ;
while(bot<top && dcmp(xmult(vec[i].s,vec[i].e,CrossPoint(deq[top],deq[top-1])))<0)
top--;//处理下凸壳
while(bot<top && dcmp(xmult(vec[i].s,vec[i].e,CrossPoint(deq[bot],deq[bot+1])))<0)
bot++;//处理上凸壳
deq[++top]=vec[i];
}
while(bot<top && dcmp(xmult(deq[bot].s,deq[bot].e,CrossPoint(deq[top],deq[top-1])))<0)
top--;//先滤除左侧的半平面。指针 p1 指向下凸壳的最左侧，p2 指向上凸壳的最左侧。

while(bot<top && dcmp(xmult(deq[top].s,deq[top].e,CrossPoint(deq[bot],deq[bot+1])))<0)
bot++;//滤除右侧没有用的半平面

if(top <= bot+1)return ;

for(m=0,i=bot;i<top;i++)
cp[m++]=CrossPoint(deq[i],deq[i+1]);
if(bot<top+1)
cp[m++]=CrossPoint(deq[bot],deq[top]);
m=unique(cp,cp+m,EqualPoint)-cp;
}

double PolygonArea(cpoint p[], int n) {
if (n < 3) return 0;
double s = p[0].y * (p[n - 1].x - p[1].x);
for (int i = 1; i < n; ++i)
s += p[i].y * (p[i - 1].x - p[(i + 1) % n].x);
return fabs(s / 2); // 顺时针方向 s为负
}
int n,m;
cvector v
;
cpoint cp
;

void solve()
{
int i,j;
setline(0, 0, 10000, 0, v[0]);//为了保证结果是凸多边形或其退化，首先建立一个非～常～大的边框
setline(10000, 0, 10000, 10000, v[1]);
setline(10000, 10000, 0, 10000, v[2]);
setline(0, 10000, 0, 0, v[3]);
cpoint p
;
double x1,y1,x2,y2;
for(j=4,i=0;i<n;i++){
scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
setline(x1,y1,x2,y2,v[j++]);
}
n=j;
HalfPanelCross(v,n,cp,m);//向量(直线)集合,长度;点集,长度
printf("%.1lf\n",PolygonArea(cp,m));
}

int main()
{
while(~scanf("%d",&n))
{
memset(cp,0,sizeof(cp));
memset(v,0,sizeof(v));
solve();
}
return 0;
}