poj2318 - TOYS

来源：互联网发布：学历网络教育报考时间编辑：程序博客网时间：2024/04/30 07:23

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题目大意：有一个矩形盒子，盒子里会有一些木块线段，并且这些线段是按照顺序给出的，有n条线段，把盒子分层了n+1个区域,然后有m个玩具，这m个玩具的坐标是已知的，问最后每个区域有多少个玩具

解题思路：因为线段是有序给出，所以不用排序，判断某个点在哪个区域，采用二分法，将某个点和线段的叉积来判断这个点是在线的左边或者右边，根据这个来二分找出区域

#include <iostream>using namespace std;#define _MAXL_ 5010typedef struct{int x, y;}POINT;typedef struct{POINT first,second;}LINE;class CBox{private:int nTop,nLeft,nRight,nButtom;int nCountLine;//记录盒子里有多少条线LINE *pLine;//记录线的信息int *pRange;//记录区间中的玩具有多少个int Multi(POINT p1,POINT p2,POINT p0);//叉积计算public:CBox();virtual ~CBox();void SetBorder(int top,int left,int right,int buttom);//设置盒子矩形坐标void AddLine(int first,int second);//在盒子中增加一条线int GetRange(int val);//得到某个区间中的玩具个数void SetToy(POINT toy);//在盒子里放一个玩具};CBox::CBox(){nCountLine = 0;pLine = new LINE[_MAXL_];pRange = new int[_MAXL_];memset(pRange,0,sizeof(pRange)*_MAXL_);}CBox::~CBox(){delete []pLine;delete []pRange;}void CBox::SetBorder(int top,int left,int right,int buttom){nTop = top;nLeft = left;nRight = right;nButtom = buttom;}void CBox::AddLine(int topx,int buttomx){pLine[nCountLine].first.x = topx;pLine[nCountLine].first.y = nTop;pLine[nCountLine].second.x = buttomx;pLine[nCountLine].second.y = nButtom;nCountLine++;}int CBox::GetRange(int val){return pRange[val];}int CBox::Multi(POINT p1,POINT p2,POINT p0){ return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);}void CBox::SetToy(POINT toy){//利用二分查找，查找出该点在哪个区间int l, r, mid;l=0,r=nCountLine-1;while (l < r){mid = (l+r)>>1;if (Multi(toy, pLine[mid].first, pLine[mid].second) > 0) l = mid + 1;else r = mid;}  if (Multi(toy, pLine[l].first, pLine[l].second) < 0) pRange[l]++;else pRange[l+1]++;}int main(){int i,fx,lx;POINT a,b;int n,m;while(cin>>n && n){cin>>m>>a.x>>a.y>>b.x>>b.y;CBox box;box.SetBorder(a.y,a.x,b.x,b.y);for(i = 0;i < n;i++){cin>>fx>>lx;box.AddLine(fx,lx);}for(i = 0;i < m;i++){cin>>a.x>>a.y;box.SetToy(a);}for(i = 0;i <= n;i++){cout<<i<<": "<<box.GetRange(i)<<endl;}cout<<endl;}return 0;}