poj 1514 Metal Cutting 半平面交的平面切割

半平面交的模板O(n^2)在http://www.cnblogs.com/wuyiqi/archive/2012/03/30/2426175.html

题意：判断在一个正方形上切出题目要求的凸多边形的最短切痕总和

总共才8条边，数据规模不大，时限10s，直接暴力枚举切割的顺序，然后计算

这是我第一次接触平面切割，也可以算是处女割了吧，呵呵

#include <cmath>
#include <cstdio>
#include<algorithm>
using namespace std;
const  int maxn = 100;
const double eps = 1e-8;
inline double min(double a,double b){    return a<b?a:b;}
inline double sgn(double x) {return fabs(x)<eps?0:(x>0?1:-1);}
struct Point{
    double x,y;
    Point(double tx=0,double ty=0){x=tx;y=ty;}
    bool operator == (const Point& t) const {
        return sgn(x-t.x)==0 && sgn(y-t.y)==0;
    }
}p[maxn],Set[maxn],st[maxn],tmp[maxn],pp[maxn];
struct Seg{Point s,e;};
inline double dist(Point a,Point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}
inline double cross(Point a,Point b,Point c){return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}
inline bool outside(Seg seg,Point p){    return cross(seg.s,seg.e,p)>eps;}
inline bool inside(Seg seg,Point p){    return cross(seg.s,seg.e,p)<-eps;}
Point Intersect(Point p1, Point p2, Point p3, Point p4, Point& p) {
    double a1, b1, c1, a2, b2, c2, d;
    a1 = p1.y - p2.y; b1 = p2.x - p1.x; c1 = p1.x*p2.y - p2.x*p1.y;
    a2 = p3.y - p4.y; b2 = p4.x - p3.x; c2 = p3.x*p4.y - p4.x*p3.y;
    d = a1*b2 - a2*b1;
    if ( fabs(d) < eps )    return false;
    p.x = (-c1*b2 + c2*b1) / d;
    p.y = (-a1*c2 + a2*c1) / d;
    return p;
}
double W,H;
int a[10],n,pn;
double CUT(Seg seg,Point p[]){
    int i,j,tot=0;
    Point A,B;
    A=B=Point(0,0);
    bool s,e;
    for(i=0;i<pn;i++){
        if(!outside(seg,p[i]))  pp[tot++]=p[i];
        else {
            if(i==0&&!outside(seg,p[pn-1])){B=A;
                pp[tot++]=Intersect(seg.s,seg.e,p[i],p[pn-1],A);
            }
            if(i!=0&&!outside(seg,p[i-1])) {B=A;
                pp[tot++]=Intersect(seg.s,seg.e,p[i],p[i-1],A);
            }
            if(!outside(seg,p[i+1])) {
                B=A;
                pp[tot++]=Intersect(seg.s,seg.e,p[i],p[i+1],A);
            }
        }
    }
        pp[tot]=pp[0];pn=tot;
        memcpy(st,pp,sizeof(pp));
    return dist(A,B);
}
int main(){
    int i;
    while(scanf("%lf%lf",&W,&H)!=EOF){
        double ans=1e20;
        st[4]=st[0]=Point(0,0);
        st[1]=Point(0,H);
        st[2]=Point(W,H);
        st[3]=Point(W,0);
        memcpy(tmp,st,sizeof(st));
        scanf("%d",&n);Seg ts[100];
        for(i=0;i<n;i++) {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            a[i]=i;
        }p[n]=p[0];
        for(i=0;i<n;i++) ts[i].s=p[i],ts[i].e=p[i+1];
        do{
            double tlen=0;
            memcpy(st,tmp,sizeof(tmp));
            pn=4;
            for(i=0;i<n;i++){
                tlen+=CUT(ts[a[i]],st);
            }
            ans=min(ans,tlen);
        }while(next_permutation(a,a+n));
        printf("Minimum total length = %.3lf\n",ans);
    }
    return 0;
}