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;
}

POJ 1279 Art Gallery 半平面交求多边形核