[USACO3.4.1 Closed Fences]

[关键字]：计算几何

[题目大意]：给出N个点，先判断这些点能否构成一个边与边不相交的多边形，然后在判断站在给出点能看见哪几条边。

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[分析]：第一问比较简单，只要判断规范相交就行了，利用叉积可以很容易的判断，祥见黑书。第二问可以用二分法解决，每次都找一条边的中点然后判断从给定点到中点是否和别的边相交，如果相交就二分判断能否看到左边或右边的部分，结束条件就是当前部分长度小于某个设定值。

[代码]：

View Code

/*
ID:procedure2
PROB:fence4 
LANG:C++
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int MAXN=300;
const double JD=0.005;

struct node
{
       double x,y;
}point[MAXN],eye;
struct rec
{
       node a,b;
}edge[MAXN];
int N,ans=0;
bool v[MAXN],b[MAXN];

double cross(node a,node b,node c)
{
    double x1=b.x-a.x,y1=b.y-a.y;
    double x2=c.x-a.x,y2=c.y-a.y;
    //printf("%0.lf %0.lf %0.lf %0.lf %0.lf %0.lf %0.lf\n",a.x,a.y,b.x,b.y,c.x,c.y,x1*y2-x2*y1);
    return x1*y2-x2*y1;
}

int Cleck(rec A,rec B)
{
     double t1=cross(A.a,A.b,B.a),t2=cross(A.a,A.b,B.b);
     double t3=cross(B.a,B.b,A.a),t4=cross(B.a,B.b,A.b);
     //printf("%lf %lf ****\n",t1,t2);
     if (t1==0 || t2==0) return -1;
     if (t1*t2<0 && t3*t4<0) return 1;
     return 0;
}

bool Judge(const rec &mid)
{
    for (int i=1;i<=N;i++)
        if (!b[i])
            if (Cleck(mid,edge[i])!=0) return 0;
    return 1;
}

bool Watch(node A,node B)
{
    if (sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y))<JD) return 0;
    rec mid;
    bool flag=0;
    mid.b.x=(A.x+B.x)/2;
    mid.b.y=(B.y+A.y)/2;
    mid.a.x=eye.x;
    mid.a.y=eye.y;
    //printf("%.0lf %.0lf %0.lf %0.lf\n",A.x,A.y,B.x,B.y);
    if (Judge(mid)) return 1;
    else
    {
        flag=Watch(A,mid.b);
        if (!flag) flag=Watch(mid.b,B);
    }
    return flag;
}

int main()
{
    freopen("fence4.in","r",stdin);
    freopen("fence4.out","w",stdout);
    scanf("%d",&N);
    scanf("%lf%lf",&eye.x,&eye.y);
    scanf("%lf%lf",&point[1].x,&point[1].y);
    for (int i=2;i<=N;i++)
    {
        scanf("%lf%lf",&point[i].x,&point[i].y);
        edge[i-1].a=point[i-1];
        edge[i-1].b=point[i];
    }
    edge[N].a=point[1];
    edge[N].b=point[N];
    /*for (int i=1;i<=M;i++)
        printf("%.0lf %.0lf %.0lf %.0lf\n",edge[i].a.x,edge[i].a.y,edge[i].b.x,edge[i].b.y);*/
    //printf("%d %d\n",N,M);
    for (int i=1;i<N;i++)
        for (int j=i+1;j<=N;j++)
            if (Cleck(edge[i],edge[j])==1) {printf("NOFENCE\n");return 0;}
    memset(b,0,sizeof(b));
    memset(v,0,sizeof(v));
    for (int i=1;i<=N;i++)
    {
        if (cross(eye,edge[i].a,edge[i].b)==0) continue;
        b[i]=1;
        if (Watch(edge[i].a,edge[i].b)) ans++,v[i]=1;
        b[i]=0;
    }
    printf("%d\n",ans);
    if (v[N] && v[N-1]) swap(edge[N],edge[N-1]);
    for (int i=1;i<=N;i++)
        if (v[i]) printf("%0.lf %0.lf %0.lf %0.lf\n",edge[i].a.x,edge[i].a.y,edge[i].b.x,edge[i].b.y);
    return 0;
}