poj 1279 Art Gallery

【題目大意】:時計回りにn個の点を与えて、多角形の核の面積を求めます
【解題の考え方】:半面交模版テスト
【コード】:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <cctype>
#include <map>
#include <iomanip>
                   
using namespace std;
                   
#define eps 1e-8
#define pi acos(-1.0)
#define inf 1<<30
#define linf 1LL<<60
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long

#define MAXN 2000

struct Point {
    double x,y;
    Point() {}
    Point(double _x,double _y){
        x=_x,y=_y;
    }
};
/* （ ）（ 1 ）*/  
Point points[MAXN],p[MAXN],q[MAXN];  
int n;  
double r;  
int cCnt,curCnt;  

int sig(double k){
    return (k<-eps)?-1:(k>eps);
}


inline void getline(Point x,Point y,double &a,double &b,double &c){  
    a=y.y-x.y;  
    b=x.x-y.x;  
    c=y.x*x.y-x.x*y.y;  
}  

inline void initial(){  
    for (int i=1; i<=n; ++i) p[i]=points[i];  
    p[n+1]=p[1];  
    p[0]=p[n];  
    cCnt=n;  
}  

inline Point intersect(Point x,Point y,double a,double b,double c){  
    double u=fabs(a*x.x+b*x.y+c);  
    double v=fabs(a*y.x+b*y.y+c);  
    return Point((x.x*v+y.x*u)/(u+v),(x.y*v+y.y*u)/(u+v));  
}  

inline void cut(double a,double b ,double c){  
    curCnt=0;  
    for (int i=1; i<=cCnt; ++i){  
        if (sig(a*p[i].x+b*p[i].y+c)>=0) q[++curCnt]=p[i];  
        else {  
            if (sig(a*p[i-1].x+b*p[i-1].y+c)>0){  
                q[++curCnt]=intersect(p[i],p[i-1],a,b,c);  
            }  
            if(sig(a*p[i+1].x+b*p[i+1].y+c)>0){  
                q[++curCnt]=intersect(p[i],p[i+1],a,b,c);  
            }  
        }  
    }  
    for (int i=1; i<=curCnt; ++i) p[i]=q[i];  
    p[curCnt+1]=p[1];
    p[0]=p[curCnt];  
    cCnt=curCnt;
    return ;  
}  

inline void solve(){  
    // ： ， ，   
    initial(); 
    for(int i=1; i<=n; ++i){  
        double a,b,c;  
        getline(points[i],points[i+1],a,b,c);  
        cut(a,b,c);  
    }  
    // 
    double area=0;  
    for (int i=1; i<=curCnt; ++i)  
      area+=p[i].x*p[i+1].y-p[i+1].x*p[i].y;  
    area=fabs(area/2.0); 
    printf("%.2f
",area); 
    // cCnt ，p   
}

inline void GuiZhengHua(){  
    // ， ，   
    for(int i = 1; i < (n+1)/2; i ++)  
      swap(points[i], points[n-i]);// iostream  
} 

int main() {
    int T;
    cin >> T;
    while (T--){
        scanf("%d",&n);
        double a,b;
        for (int i=1; i<=n; i++){
            scanf("%lf%lf",&a,&b);
            points[i]=Point(a,b);
        }
        points[n+1]=points[1];
        //  GuiZhengHua();
        solve();
    }
    return 0;
}