poj 2074

哎怎么说，感觉现在处理平面上点线的题已经比较熟练了。

这题就离散化然后搞个前缀和就没了。

准备开始进一步的自闭了。

下面是disguss的一些样例。。。

其实是我自己写错了个地方，本来能1A的。

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <iostream>
 #include <iomanip>
 #include <vector>
 #include <algorithm>
 using namespace std;
 typedef double db;
 const db eps=1e-;
 const db pi=acos(-);
 int sign(db k){
     if (k>eps) return ; else if (k<-eps) return -; return ;
 }
 int cmp(db k1,db k2){return sign(k1-k2);}
 struct point{
     db x,y;
     point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
     point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
     point operator * (db k1) const{return (point){x*k1,y*k1};}
     point operator / (db k1) const{return (point){x/k1,y/k1};}
     db abs(){ return sqrt(x*x+y*y);}
 };
 db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
 db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
 point getLL(point k1,point k2,point k3,point k4){
     db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
 }
 int intersect(db l1,db r1,db l2,db r2){
     if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-&&cmp(r2,l1)!=-;
 }
 int checkSS(point k1,point k2,point k3,point k4){
     return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
            sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=&&
            sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=;
 }
 struct line{
     point p[];
     line(point k1,point k2){p[]=k1; p[]=k2;}
 };
 double l,r,y;
 point p[];
 int n;
 point s1,t1,s2,t2;
 //离散化+前缀和。
 vector<db> v,a,b;
 int pre[];
 int main(){
     ios::sync_with_stdio(false);
     cout<<fixed<<setprecision();
     while (cin>>l>>r>>y&&(l||r||y)){
         s1={l,y};t1={r,y};
         cin>>l>>r>>y;
         s2={l,y};t2={r,y};
         v.push_back(l);
         v.push_back(r);
         cin>>n;
         for(int i=;i<=*n;i+=){
             cin>>l>>r>>y;
             p[i]={l,y};
             p[i+]={r,y};
         }
         for(int i=;i<=*n;i+=){
             if(cmp(s1.y,p[i].y)==&&cmp(p[i].y,s2.y)==){//在中间
                 point tmp = getLL(t1,p[i],s2,t2);
                 v.push_back(tmp.x);//-1
                 a.push_back(tmp.x);
                 tmp = getLL(s1,p[i+],s2,t2);
                 v.push_back(tmp.x);//+1
                 b.push_back(tmp.x);
             }
         }
         sort(v.begin(),v.end());
         v.erase(unique(v.begin(),v.end()),v.end());
         for(int i=;i<a.size();i++){
             int id = lower_bound(v.begin(),v.end(),a[i])-v.begin();
             pre[id]--;
         }
         for(int i=;i<b.size();i++){
             int id = lower_bound(v.begin(),v.end(),b[i])-v.begin();
             pre[id]++;
         }
         int id1 = lower_bound(v.begin(),v.end(),s2.x)-v.begin();
         int id2 = lower_bound(v.begin(),v.end(),t2.x)-v.begin();
         for(int i=;i<v.size();i++){
             pre[i]+=pre[i-];
         }
         db ans = ;
         for(int l=id1,r;l<=id2;l=r+){
             r=l;
             while (r<id2&&pre[r]>=)
                 r++;
             ans=max(ans,v[r]-v[l]);
         }
         if(cmp(ans,0.00)==){
             cout<<"No View"<<endl;
         } else {
             cout << ans << endl;
         }
         v.clear();
         a.clear();
         b.clear();
         memset(pre,, sizeof(pre));
     }
 }
 /**
  2 6 6
  0 15 0
  1
  2 6 7

  2 6 6
  0 15 0
  1
  2 6 7
  2 6 6
  0 15 0
  1
  4 4.5 5.5
  2 6 6
  0 15 0
  16
  0 1 3
  1.5 2 3
  2.5 3 3
  3.5 4 3
  4.5 5 3
  5.5 6 3
  6.5 7 3
  7.5 8 3
  8.5 9 3
  9.5 10 3
  10.5 11 3
  11.5 12 3
  12.5 13 3
  13.5 14 3
  14.5 15 3
  15.5 16 3
  2 6 6
  0 15 0
  16
  0 1 .1
  1.5 2 .1
  2.5 3 .1
  3.5 4 .1
  4.5 5 .1
  5.5 6 .1
  6.5 7 .1
  7.5 8 .1
  8.5 9 .1
  9.5 10 .1
  10.5 11 .1
  11.5 12 .1
  12.5 13 .1
  13.5 14 .1
  14.5 15 .1
  15.5 16 .1
  2 6 6
  0 15 0
  14
  0 1 3
  1.5 2 3
  2.5 3 3
  3.5 4 3
  4.5 5 3
  5.5 6 3
  8.5 9 3
  9.5 10 3
  10.5 11 3
  11.5 12 3
  12.5 13 3
  13.5 14 3
  14.5 15 3
  15.5 16 3

  2 6 6
  0 15 1
  5
  1 1.5 6
  17 18 1
  3 5 3
  0 20 10
  0 20 0.5
  */