POJ2451 Uyuw's Concert

听大佬说这题要用O(nlogn)的半平面交过，于是我。。。。我就继续使用我的O(nlogn)的板子，哈哈哈哈哈嗝

板子用的kuangbin大佬的http://www.cnblogs.com/kuangbin/p/3266097.html

题意：在（0，10000）*（0，10000）的坐标系上，给定n个半平面，求出它们围成的图形的面积

把坐标边界线加进去，叉积求面积就OK 啦~~

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include<cmath>using namespace std;const double eps = 1e-8;const double PI = acos(-1.0);int sgn(double x){    if(fabs(x) < eps) return 0;    if(x < 0) return -1;    else return 1;}struct Point{    double x,y;    Point(){}    Point(double _x,double _y)    {        x = _x; y = _y;    }    Point operator -(const Point &b)const    {        return Point(x - b.x, y - b.y);    }    double operator ^(const Point &b)const    {        return x*b.y - y*b.x;    }    double operator *(const Point &b)const    {        return x*b.x + y*b.y;    }};struct Line{    Point s,e;    double k;    Line(){}    Line(Point _s,Point _e)    {        s = _s; e = _e;        k = atan2((e.y - s.y),(e.x - s.x));    }    Point operator &(const Line &b)const    {        Point res = s;        double t = ((s - b.s)^(b.s - b.e))/((s - e)^(b.s - b.e));        res.x += (e.x - s.x)*t;        res.y += (e.y - s.y)*t;        return res;    }};//半平面交，直线的左边代表有效区域bool HPIcmp(Line a,Line b){    if(fabs(a.k - b.k) > eps)return a.k < b.k;    return ((a.s - b.s)^(b.e - b.s)) < 0;}Line Q[20005];void HPI(Line line[], int n, Point res[], int &resn){    int tot = n;    sort(line,line+n,HPIcmp);    tot = 1;    for(int i = 1;i < n;i++)        if(fabs(line[i].k - line[i-1].k) > eps)            line[tot++] = line[i];    int head = 0, tail = 1;    Q[0] = line[0];    Q[1] = line[1];    resn = 0;    for(int i = 2; i < tot; i++)    {        if(fabs((Q[tail].e-Q[tail].s)^(Q[tail-1].e-Q[tail-1].s)) < eps || fabs((Q[head].e-Q[head].s)^(Q[head+1].e-Q[head+1].s)) < eps)            return;        while(head < tail && (((Q[tail]&Q[tail-1]) - line[i].s)^(line[i].e-line[i].s)) > eps)            tail--;        while(head < tail && (((Q[head]&Q[head+1]) - line[i].s)^(line[i].e-line[i].s)) > eps)            head++;        Q[++tail] = line[i];    }    while(head < tail && (((Q[tail]&Q[tail-1]) - Q[head].s)^(Q[head].e-Q[head].s)) > eps)        tail--;    while(head < tail && (((Q[head]&Q[head-1]) - Q[tail].s)^(Q[tail].e-Q[tail].e)) > eps)        head++;    if(tail <= head + 1)return;    for(int i = head; i < tail; i++)        res[resn++] = Q[i]&Q[i+1];    if(head < tail - 1)        res[resn++] = Q[head]&Q[tail];}Point p[20005];//p数组存结果Line line[20005];double xmult(Point p0,Point p1,Point p2){/*p0p1 X p0p2*/    return (p1-p0)^(p2-p0);}int main(){    int n;    while(~scanf("%d",&n))    {        for(int i = 0;i <n;i++){            scanf("%lf%lf",&p[i].x,&p[i].y);            scanf("%lf%lf",&p[i+1].x,&p[i+1].y);            line[i] = Line(p[i],p[i+1]);        }        line[n++]=Line( Point(0, 0), Point(10000, 0));        line[n++]=Line( Point(10000, 0), Point(10000, 10000));        line[n++]=Line( Point(10000, 10000), Point(0, 10000));        line[n++]=Line( Point(0, 10000), Point(0, 0));        int resn;        double area=0.0;        HPI(line,n,p,resn);        for(int i=1;i<resn-1;i++)            area+=xmult(p[0],p[i],p[i+1]);        if(resn<3) area=0;        if(area<0) area=-area;        printf("%.1f\n",area/2.0);    }    return 0;}