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。
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <iostream> 5 #include <iomanip> 6 #include <vector> 7 #include <algorithm> 8 using namespace std; 9 typedef double db; 10 const db eps=1e-6; 11 const db pi=acos(-1); 12 int sign(db k){ 13 if (k>eps) return 1; else if (k<-eps) return -1; return 0; 14 } 15 int cmp(db k1,db k2){return sign(k1-k2);} 16 struct point{ 17 db x,y; 18 point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};} 19 point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};} 20 point operator * (db k1) const{return (point){x*k1,y*k1};} 21 point operator / (db k1) const{return (point){x/k1,y/k1};} 22 db abs(){ return sqrt(x*x+y*y);} 23 }; 24 db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;} 25 db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;} 26 point getLL(point k1,point k2,point k3,point k4){ 27 db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2); 28 } 29 int intersect(db l1,db r1,db l2,db r2){ 30 if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1; 31 } 32 int checkSS(point k1,point k2,point k3,point k4){ 33 return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&& 34 sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&& 35 sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0; 36 } 37 struct line{ 38 point p[2]; 39 line(point k1,point k2){p[0]=k1; p[1]=k2;} 40 }; 41 double l,r,y; 42 point p[10005]; 43 int n; 44 point s1,t1,s2,t2; 45 //离散化+前缀和。 46 vector<db> v,a,b; 47 int pre[100005]; 48 int main(){ 49 ios::sync_with_stdio(false); 50 cout<<fixed<<setprecision(2); 51 while (cin>>l>>r>>y&&(l||r||y)){ 52 s1={l,y};t1={r,y}; 53 cin>>l>>r>>y; 54 s2={l,y};t2={r,y}; 55 v.push_back(l); 56 v.push_back(r); 57 cin>>n; 58 for(int i=1;i<=2*n;i+=2){ 59 cin>>l>>r>>y; 60 p[i]={l,y}; 61 p[i+1]={r,y}; 62 } 63 for(int i=1;i<=2*n;i+=2){ 64 if(cmp(s1.y,p[i].y)==1&&cmp(p[i].y,s2.y)==1){//在中间 65 point tmp = getLL(t1,p[i],s2,t2); 66 v.push_back(tmp.x);//-1 67 a.push_back(tmp.x); 68 tmp = getLL(s1,p[i+1],s2,t2); 69 v.push_back(tmp.x);//+1 70 b.push_back(tmp.x); 71 } 72 } 73 sort(v.begin(),v.end()); 74 v.erase(unique(v.begin(),v.end()),v.end()); 75 for(int i=0;i<a.size();i++){ 76 int id = lower_bound(v.begin(),v.end(),a[i])-v.begin(); 77 pre[id]--; 78 } 79 for(int i=0;i<b.size();i++){ 80 int id = lower_bound(v.begin(),v.end(),b[i])-v.begin(); 81 pre[id]++; 82 } 83 int id1 = lower_bound(v.begin(),v.end(),s2.x)-v.begin(); 84 int id2 = lower_bound(v.begin(),v.end(),t2.x)-v.begin(); 85 for(int i=1;i<v.size();i++){ 86 pre[i]+=pre[i-1]; 87 } 88 db ans = 0; 89 for(int l=id1,r;l<=id2;l=r+1){ 90 r=l; 91 while (r<id2&&pre[r]>=0) 92 r++; 93 ans=max(ans,v[r]-v[l]); 94 } 95 if(cmp(ans,0.00)==0){ 96 cout<<"No View"<<endl; 97 } else { 98 cout << ans << endl; 99 } 100 v.clear(); 101 a.clear(); 102 b.clear(); 103 memset(pre,0, sizeof(pre)); 104 } 105 } 106 /** 107 2 6 6 108 0 15 0 109 1 110 2 6 7 111 112 2 6 6 113 0 15 0 114 1 115 2 6 7 116 2 6 6 117 0 15 0 118 1 119 4 4.5 5.5 120 2 6 6 121 0 15 0 122 16 123 0 1 3 124 1.5 2 3 125 2.5 3 3 126 3.5 4 3 127 4.5 5 3 128 5.5 6 3 129 6.5 7 3 130 7.5 8 3 131 8.5 9 3 132 9.5 10 3 133 10.5 11 3 134 11.5 12 3 135 12.5 13 3 136 13.5 14 3 137 14.5 15 3 138 15.5 16 3 139 2 6 6 140 0 15 0 141 16 142 0 1 .1 143 1.5 2 .1 144 2.5 3 .1 145 3.5 4 .1 146 4.5 5 .1 147 5.5 6 .1 148 6.5 7 .1 149 7.5 8 .1 150 8.5 9 .1 151 9.5 10 .1 152 10.5 11 .1 153 11.5 12 .1 154 12.5 13 .1 155 13.5 14 .1 156 14.5 15 .1 157 15.5 16 .1 158 2 6 6 159 0 15 0 160 14 161 0 1 3 162 1.5 2 3 163 2.5 3 3 164 3.5 4 3 165 4.5 5 3 166 5.5 6 3 167 8.5 9 3 168 9.5 10 3 169 10.5 11 3 170 11.5 12 3 171 12.5 13 3 172 13.5 14 3 173 14.5 15 3 174 15.5 16 3 175 176 177 2 6 6 178 0 15 1 179 5 180 1 1.5 6 181 17 18 1 182 3 5 3 183 0 20 10 184 0 20 0.5 185 */