USACO 3.4.1 closed fence

       不得不承认这道题着实是一道好题，可惜当年我还在用PASCAL的时候压根没做这道题……具体详情我想我是需要追究matt和s.p.willam（原skyprophet）两位神牛的责任的……

      这个东西比较困难，我也不知道完美算法是什么（或者压根没有完美算法？？？），于是我祭出了近似算法。这道题目的第一问就是简单的判断线段是否相交，非常简单。由于n最大是200，所以就不用考虑O(nlogn)的扫除线法了——否则我就不会啦！第二问很有意思，问你能看到哪些线段。我们想：如果要看到某条线度，一定是通过某两个顶点之间的夹缝看到的。当然对于直接看到的线段，目光经过的这个夹缝可以说是很大很大，但是我们可以抽象地确定它的存在。这样我们只需要搞出来这些“夹缝”！我的想法是近似的让目光游离于各个顶点的四周，也就是将各个顶点想四周移动后引射线作为目光，之后寻找具光源最近的交点所在线段，这条线段就是能被看到的之一！

 

/*ID: wangqia6TASK: fence4LANG: C++*/#include <fstream>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double da = 0.0000123456;const double db = 0.0000456789;const double zero = 1e-6;const double inf = 20000000;const long NMAX = 205;struct point{ double x,y;};ifstream fin ("fence4.in");ofstream fout ("fence4.out");long n,tot = 0;point o,p[NMAX],tmp;bool vis[NMAX];void initdata(){ fin >> n; fin >> o.x >> o.y; for (long i = 1; i <=n; i++) fin >> p[i].x >> p[i].y; p[n + 1] = p[1]; return; }double det(double xa,double ya,double xb,double yb){ return xa * yb - ya * xb; }double cross(point pa,point pb,point pc){ return det(pb.x - pa.x,pb.y - pa.y,pc.x - pa.x,pc.y - pa.y); }bool diff(double da,double db){ bool flag = (da > zero && db < -zero) || (da < -zero && db > zero); return flag;}bool eql_zero(double x){ return x <zero && x > -zero;}bool on_seg(point pa,point pb,point pc){ return min(pa.x,pb.x) <= pc.x && max(pa.x,pb.x) >= pc.x && min(pa.y,pb.y) <= pc.y && max(pa.y,pb.y) >= pc.y;}bool intersect(point pa,point pb,point pc,point pd){ double da = cross(pa,pb,pc), db = cross(pa,pb,pd), dc = cross(pc,pd,pa), dd = cross(pc,pd,pb); if (diff(da,db) && diff(dc,dd)) return true; else return (eql_zero(da) && on_seg(pa,pb,pc)) || (eql_zero(db) && on_seg(pa,pb,pd)) || (eql_zero(dc) && on_seg(pc,pd,pa)) || (eql_zero(dd) && on_seg(pc,pd,pb));}bool worka(){ long i,j; for (i = 1; i < n - 1; i++) for (j = i + 2; j <= n; j++) if (intersect(p[i],p[i + 1],p[j],p[j + 1]) && !((i == 1) && (j == n))) return true; return false; }point find_point(point pa,point pb,point pc,point pd){ point k; double da = cross(pa,pb,pc),db = cross(pa,pb,pd); k.x = (da * pd.x - db * pc.x) / (da - db); k.y = (da * pd.y - db * pc.y) / (da - db); return k; }double sqr(double x){ return x * x; }double dist(point a,point b){ return sqr(a.x - b.x) + sqr(a.y - b.y); }void lookup(){ point k; double min_tmp = inf,dist_tmp; long min_vex = 0; for (long i = 1; i <= n; i++) if (intersect(o,tmp,p[i],p[i + 1])) { k = find_point(o,tmp,p[i],p[i + 1]); dist_tmp = dist(o,k); if (dist_tmp + zero < min_tmp) { min_tmp = dist_tmp; min_vex = i; } } if ((min_vex > 0) && (! vis[min_vex])) { tot++; vis[min_vex] = true; }}void workb(){ memset(vis,0,sizeof(vis)); for (long i = 1; i <= n; i++) { tmp.x = (p[i].x + da - o.x) * inf; tmp.y = (p[i].y + db - o.y) * inf; lookup(); tmp.x = (p[i].x - da - o.x) * inf; tmp.y = (p[i].y - db - o.y) * inf; lookup(); tmp.x = (p[i].x + da - o.x) * inf; tmp.y = (p[i].y - db - o.y) * inf; lookup(); tmp.x = (p[i].x - da - o.x) * inf; tmp.y = (p[i].y + db - o.y) * inf; lookup(); } }void write(point pa,point pb){ fout << pa.x << ' ' << pa.y << ' ' << pb.x << ' ' << pb.y << endl; }void outitdata(){ fout << tot << endl; for (long i = 1; i < n - 1; i++) if (vis[i]) write(p[i],p[i + 1]); if (vis[n]) write(p[1],p[n]); if (vis[n - 1]) write(p[n - 1],p[n]); return; }int main(){ initdata(); if (worka()) { fout << "NOFENCE" << endl; fin.close(); fout.close(); return 0; } workb(); outitdata(); fin.close(); fout.close(); return 0; }