# BZOJ 1091([SCOI2003]分割多边形-分割直线)

## 1091: [SCOI2003]分割多边形

Time Limit: 1 Sec  Memory Limit: 162 MB
Submit: 223  Solved: 82
[Submit][Status]

100 100
4
80 80
70 30
20 20
20 80

312.575

## Source

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MP make_pair
#define MAXP (500+10)
#define MAXN (500+10)
#define MAXM (500+10)
#define eps (1e-6)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,m,p;
double sqr(double x){return x*x;}
int dcmp(double a,double b=0){if (fabs(a-b)<=eps) return 0;else if (a<b) return -1;return 1;}
struct P
{
double x,y;
P(){}
P(double _x,double _y):x(_x),y(_y){}
friend istream& operator>>(istream& cin,P &a){cin>>a.x>>a.y;return cin;}
friend ostream& operator<<(ostream& cout,P &a){cout<<a.x<<' '<<a.y;return cout;}
friend bool operator==(P a,P b){return dcmp(a.x,b.x)==0&&dcmp(a.y,b.y)==0;    }
}a[MAXP];
struct V
{
double x,y;
V(){}
V(double _x,double _y):x(_x),y(_y){}
V(P a,P b):x(b.x-a.x),y(b.y-a.y){}
friend V operator*(double a,V b){return V(a*b.x,a*b.y);}
friend V operator-(P a,P b){return V(b.x-a.x,b.y-a.y); }
friend double operator*(V a,V b){return a.x*b.y-a.y*b.x;}
friend double operator^(V a,V b){return a.x*b.x+a.y*b.y;}
friend P operator+(P a,V b){return P(a.x+b.x,a.y+b.y);    }
friend double dis2(V a){return sqr(a.x)+sqr(a.y);    }
};
struct L
{
P p;
V v;
L(){}
L(P _A,V _B):p(_A),v(_B){}
friend bool parallel(L a,L b) {return (dcmp(a.v.x*b.v.y,a.v.y*b.v.x))==0;}
friend P intersect(L a,L b) //直线交点
{
V &v=a.v,&w=b.v,u=V(b.p,a.p);
double t=(w*u)/(v*w);
P c=a.p+t*v; return c;
}
friend bool inleft(P a,L b){return dcmp(b.v*V(b.p,a))>=0;    }
void print(){cout<<p.x<<' '<<p.y<<' '<<v.x<<' '<<v.y<<endl;    }
}l[MAXP],lrec[4];
bool inrec(P a){return (dcmp(a.x)>=0&&dcmp(a.x,n)<=0&&dcmp(a.y)>=0&&dcmp(a.y,m)<=0);}
L through_rec_line(L l)
{
int siz=0;P st[3];
if (dcmp(l.v.x)==0) return L(P(l.p.x,l.v.y>0?0:m),V(0,(l.v.y>0?1:-1)*m));
if (dcmp(l.v.y)==0) return L(P(l.v.x>0?0:n,l.p.y),V((l.v.x>0?1:-1)*n,0)); //至此保证不平行坐标系
Rep(i,4)
{
if (parallel(lrec[i],l)) continue;
st[++siz]=intersect(lrec[i],l);
if (!inrec(st[siz])) siz--;
if (siz==2)
{
if (st[1]==st[2]) siz--;
else
{
V a=V(st[1],st[2]);
if (dcmp(a^l.v)<0) return L(st[2],V(st[2],st[1]));
return L(st[1],a);
}
}
}
}
bool b[MAXP]={0};
double ans=1e300;
int cut_list[MAXN];
void dfs(double tot,int siz)
{
if (tot>ans) return;
if (siz==p)
{
//  Rep(i,siz) cout<<cut_list[i]<<' ';printf("%.3lf\n",ans);
ans=min(ans,tot);
return;
}

/*
if (siz==2)
{
if (cut_list[0]==2&&cut_list[1]==1)
{
cout<<' ';
}
}*/

For(i,p)
if (!b[i])
{
L x=through_rec_line(l[i]);
For(j,p)
if (!parallel(l[j],x)&&b[j])
{
P p=intersect(x,l[j]);
if (dcmp(V(p,x.p)^V(p,x.p+x.v))<0)
{
if (!inleft(x.p,l[j])) x=L(p,V(p,x.p+x.v));
else if (!inleft(x.p+x.v,l[j])) x=L(x.p,V(x.p,p));
}
}
b[i]=1;
cut_list[siz]=i;
/*
Rep(j,siz) printf("\t");
cout<<i<<endl;
printf("%.3lf\n",tot+sqrt(dis2(x.v)));
*/
dfs(tot+sqrt(dis2(x.v)),siz+1);
b[i]=0;
}
}
int main()
{
//  freopen("bzoj1091.in","r",stdin);
//  freopen("bzoj1091.out","w",stdout);

cin>>n>>m>>p;
ForD(i,p) cin>>a[i];memcpy(a+p+1,a+1,sizeof(P)*p);
For(i,p) l[i]=L(a[i],V(a[i],a[i+1]));memcpy(l+p+1,l+1,sizeof(L)*p);

lrec[0]=L(P(0,0),V(n,0)),lrec[1]=L(P(n,0),V(0,m)),lrec[2]=L(P(n,m),V(-n,0)),lrec[3]=L(P(0,m),V(0,-m));
For(i,p) l[i]=through_rec_line(l[i]);

//  For(i,p) l[i].print();

dfs(0,0);

printf("%.3lf\n",ans);
return 0;
}

posted @ 2017-05-11 20:41  gccbuaa  阅读(291)  评论(0编辑  收藏  举报