Description
Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys. John’s parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box. For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.Input
The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The following n lines contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.Output
The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.Sample Input
5 6 0 10 60 0 3 1 4 3 6 8 10 10 15 30 1 5 2 1 2 8 5 5 40 10 7 9 4 10 0 10 100 0 20 20 40 40 60 60 80 80 5 10 15 10 25 10 35 10 45 10 55 10 65 10 75 10 85 10 95 10 0Sample Output
0: 2 1: 1 2: 1 3: 1 4: 0 5: 1 0: 2 1: 2 2: 2 3: 2 4: 2 这道题讨论组都是暴力啊。。 不过我用的是二分，用叉积的性质判断这个点是在这条边的左边还是右边。 然后统计出来就好了，不过计算几何还是oop好写。。 #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; #include<vector> #include<cmath> const int maxn=5005; const double eps=1e-7; struct Point { double x,y; Point(double _x=0.00,double _y=0.00):x(_x),y(_y) {} Point operator-(const Point _T1)const { return Point(x-_T1.x,y-_T1.y); } double operator^(const Point _T1)const{ return x*_T1.y-y*_T1.x; } }; struct Seg { Point left,right; Seg(double _p1=0.00,double _p2=0.00,double _p3=0.000,double _p4=0.000){ left=Point(_p1,_p2),right=Point(_p3,_p4); } }; int n,m; Seg boundSeg,partSeg[maxn]; Point toyPoint[maxn]; int number[maxn]; double Cross(Point _t1,Point _t2,Point _t3){ return (_t2-_t1)^(_t3-_t1); } int Sign(double doubleNum){ if(doubleNum>=eps)return 1; else if(fabs(doubleNum)<eps)return 0; else return -1; } int MybinarySearch(int fi) { int left=0,right=n,mid; Seg tempSeg; Point P; int ans; while(left<=right) { mid=(left+right)>>1; double res=Cross(toyPoint[fi],partSeg[mid].left,partSeg[mid].right); if(Sign(res)>0) left=mid+1; else right=mid-1,ans =mid; } return left; } int main() { #ifdef tangge freopen("2318.txt","r",stdin); #endif // tangge bool Jendl=false; while(~scanf("%d",&n),n) { scanf("%d%lf%lf%lf%lf",&m,&boundSeg.left.x,&boundSeg.left.y,&boundSeg.right.x,&boundSeg.right.y); for(int i=0; i<n; ++i) { scanf("%lf%lf",&partSeg[i].left.x,&partSeg[i].right.x); partSeg[i].left.y=boundSeg.left.y; partSeg[i].right.y=boundSeg.right.y; } partSeg[n]=Seg(boundSeg.right.x,boundSeg.right.x,boundSeg.right.x,boundSeg.right.y); for(int i=0; i<m; ++i) { scanf("%lf%lf",&toyPoint[i].x,&toyPoint[i].y); } if(Jendl)putchar(10); Jendl=true; memset(number,0,sizeof(number)); for(int i=0; i<m; ++i) { ++number[MybinarySearch(i)]; } for(int i=0; i<=n; ++i) { printf("%d: %d\n",i,number[i]); } } return 0; }