POJ 1279 Art Gallery 半平面交求多边形核

第一道半平面交，只会写N^2。

将每条边化作一个不等式，ax+by+c>0，所以要固定顺序，方便求解。

半平面交其实就是对一系列的不等式组进行求解可行解。

如果某点在直线右侧，说明那个点在区域内，否则出现在左边，就可能会有交点，将交点求出加入。

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <cstring>

#include <cmath>

#include <stack>

#include <queue>

#include <vector>

#include <algorithm>

#define ll long long

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

using namespace std;



const int MAXN = 1555;

const double eps = 1e-8;



struct POINT{

    double x;

    double y;

    POINT() : x(0), y(0) {};

    POINT(double _x_, double _y_) : x(_x_), y(_y_) {};

};



struct LINE{

    POINT a;

    POINT b;

    LINE() {};

    LINE(POINT _a_, POINT _b_) : a(_a_), b(_b_) {};

};



POINT point[MAXN];//记录最开始的多边形

POINT temp[MAXN]; //临时保存新切割的多边形

POINT ans[MAXN]; //保存新切割出的多边形

LINE lline;

int n,m;//n的原先的点数,m是新切割出的多边形的点数



void Coefficient(const LINE & L, double & A, double & B, double & C){

    A = L.b.y - L.a.y;

    B = L.a.x - L.b.x;

    C = L.b.x * L.a.y - L.a.x * L.b.y;

}



double Cross(const POINT & a, const POINT & b, const POINT &o){

    return (a.x - o.x) * (b.y - o.y) - (b.x - o.x) * (a.y - o.y);

}



POINT Intersection(const LINE & A, const LINE & B){

    double A1, B1, C1;

    double A2, B2, C2;

    Coefficient(A, A1, B1, C1);

    Coefficient(B, A2, B2, C2);

    POINT temp_point(0, 0);

    temp_point.x = -(B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);

    temp_point.y =  (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);

    return temp_point;

}



//求面积，正为顺时针，和叉积写法有关

double PointArea(POINT p[],int n){

    double area = 0;

    for(int i = 2; i < n; ++i)

        area += Cross(p[1], p[i], p[i+1]);

    return -area / 2.0;

}



void Cut(){  //用直线ax+by+c==0切割多边形

    int cut_m = 0, i;

    double a, b, c;

    Coefficient(lline, a, b, c);

    for(i = 1; i <= m; ++i){

        if(a * ans[i].x + b*ans[i].y + c >= 0)  //题目是顺时钟给出点的,所以一个点在直线右边的话，那么带入值就会大于等于0

            temp[++cut_m] = ans[i];         //说明这个点还在切割后的多边形内，将其保留

        else{

            if(a * ans[i - 1].x + b * ans[i - 1].y + c > 0){   //该点不在多边形内，但是它和它相邻的点构成直线与

                LINE line1(ans[i - 1], ans[i]); //ax+by+c==0所构成的交点可能在新切割出的多边形内，

                temp[++cut_m] = Intersection(lline, line1); //所以保留交点

            }

            if(a * ans[i + 1].x + b * ans[i + 1].y + c > 0){

                LINE line1(ans[i + 1], ans[i]);

                temp[++cut_m] = Intersection(lline, line1); //所以保留交点

            }

        }

    }

    for(i = 1; i <= cut_m; ++i) ans[i] = temp[i];

    ans[cut_m + 1] = temp[1];

    ans[0] = temp[cut_m];

    m = cut_m;

}





void solve(){

    int i;

    point[0] = point[n];

    point[n+1] = point[1];

    for(i = 0; i <= n + 1; ++i){

        ans[i] = point[i];

    }

    m = n;

    for(i = 1;i <= n; ++i){

        lline.a = point[i];

        lline.b = point[i + 1]; //根据point[i]和point[i+1]确定直线ax+by+c==0

        Cut();  //用直线ax+by+c==0切割多边形

    }

    printf("%.2f\n",Abs(PointArea(ans,m)));

}



int main(){

    int caseNum,i;

    scanf("%d",&caseNum);

    while(caseNum--){

        scanf("%d",&n);

        for(i = 1; i <= n; ++i){

            scanf("%lf%lf",&point[i].x,&point[i].y);

        }

        solve();

    }

    return 0;

}