codeforces 498A crazy town

最新推荐文章于 2021-08-03 18:05:50 发布

baker00

最新推荐文章于 2021-08-03 18:05:50 发布

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在平面被无限直线划分的CrazyTown中，确定从家到大学所在区域的最少穿越次数。通过判断直线两侧的区域来计算最短路径。

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Crazy Town is a plane on which there are n infinite line roads. Each road is defined by the equation aix + biy + ci = 0, where ai and biare not both equal to the zero. The roads divide the plane into connected regions, possibly of infinite space. Let's call each such region a block. We define an intersection as the point where at least two different roads intersect.

Your home is located in one of the blocks. Today you need to get to the University, also located in some block. In one step you can move from one block to another, if the length of their common border is nonzero (in particular, this means that if the blocks are adjacent to one intersection, but have no shared nonzero boundary segment, then it are not allowed to move from one to another one in one step).

Determine what is the minimum number of steps you have to perform to get to the block containing the university. It is guaranteed that neither your home nor the university is located on the road.

Input

The first line contains two space-separated integers x1, y1 ( - 106 ≤ x1, y1 ≤ 106) — the coordinates of your home.

The second line contains two integers separated by a space x2, y2 ( - 106 ≤ x2, y2 ≤ 106) — the coordinates of the university you are studying at.

The third line contains an integer n (1 ≤ n ≤ 300) — the number of roads in the city. The following n lines contain 3 space-separated integers ( - 106 ≤ ai, bi, ci ≤ 106; |ai| + |bi| > 0) — the coefficients of the line aix + biy + ci = 0, defining the i-th road. It is guaranteed that no two roads are the same. In addition, neither your home nor the university lie on the road (i.e. they do not belong to any one of the lines).

Output

Output the answer to the problem.

主要就是两个思考地方，要走的步就是有多少线在两点之间，把点带入的值一正一负就行。然后就是就算是long long 一起乘也要爆啊

#include<iostream>
typedef long long LL;
using namespace std;

int main()
{
 LL a,b,c,d;
 LL n;
 cin>>a>>b;
 cin>>c>>d;
 cin>>n;
 LL cross=0;
 LL x,y,z;
 for(LL i=0;i<n;i++)
 {
  cin>>x>>y>>z;
  //if((a*x+b*y+z)*(c*x+d*y+z)<0)cross++;//两个longlong要乘爆啊 
  if((a*x+b*y+z)>0&&(c*x+d*y+z)<0)cross++;
  if((a*x+b*y+z)<0&&(c*x+d*y+z)>0)cross++;
 }
 cout<<cross<<endl;
 return 0;
}