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UVA___10535
The shooter is in a great problem. He is trapped in a “2D” maze with a laser gun and can use it once. The gun is very powerful and the laser ray, it emanates can traverse infinite distance in its direction. In the maze the targets are some walls (Here this is line segments). If the laser ray touches any wall or intersects it, that wall will be destroyed and the ray will advance ahead. The shooter wants to know the maximum number of walls, he can destroy with a single shot. The shooter will never stand on a wall.
Input
The input file contains 100 sets which needs to be processed. The description of each set is given below: Each set starts with a postive integer, N (1 ≤ N ≤ 500) the number of walls. In next few lines there will be 4 ∗ N integers indicating two endpoints of a wall in cartesian co-ordinate system. Next line will contain (x, y) the coordinates of the shooter. All coordinates will be in the range [-10000,10000]. Input is terminated by a case where N = 0. This case should not be processed.
Output
For each set of input print the maximum number of walls, he can destroy by a single shot with his gun in a single line.
Sample Input
3
0 0 10 0
0 1 10 1
0 2 10 2
0 -1
3
0 0 10 0
0 1 10 1
0 3 10 3
0 2
0
Sample Output
3
2
题意：
从原点开枪，枪的子弹可以穿透墙壁，给出所有墙壁的两个端点以及原点，求最多可以穿透多少墙壁（只能朝一个方向开一枪）
思路：
可以将墙与原点的关系转化为可达角度，然后就会形成一个区间，那么问题模型就变为选择一个点，求覆盖到它的最多的区间
#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> using namespace std; #define PI acos(-1.0) struct node { double r; int type; bool operator < (const struct node&another) const { if(r!=another.r) return r<another.r; return type<another.type; } }event[5510]; struct cc { int x0,y0,x1,y1; }contra[510]; int x,y; double caculate(int x0,int y0) { if(x0==x&&y0>y) return 90; if(x0==x&&y0<y) return 270; if(x0<x&&y0==y) return 180; if(x0>x&&y0==y) return 0; double acor=atan((y0-y)*1.0/(x0-x)); if(acor<0) acor+=PI; if(y0<y) return 180+acor*180/(PI); return acor*180/PI; } int main() { int n; while(~scanf("%d",&n),n!=0) { for(int i=0;i<n;i++) cin>>contra[i].x0>>contra[i].y0>>contra[i].x1>>contra[i].y1; cin>>x>>y; int id=0; for(int i=0;i<n;i++) { int l=caculate(contra[i].x0,contra[i].y0); int r=caculate(contra[i].x1,contra[i].y1); if(l>r) swap(l,r); if(r-l>=180) { event[id].r=0;event[id++].type=-1; event[id].r=l;event[id++].type=1; event[id].r=r;event[id++].type=-1; event[id].r=360;event[id++].type=1; continue; } event[id].r=l;event[id++].type=-1; event[id].r=r;event[id++].type=1; } sort(event,event+id); int maxx=0; int cur=0; for(int i=0;i<id;i++) { if(event[i].type==-1) {cur++;maxx=max(maxx,cur);} else cur--; } cout<<maxx<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/zsyacm666666/p/4909021.html
