TOYS(计算几何-入门)

题目

‘^’代表叉乘

‘?’代表点乘

点积：a?b=ax*bx+ay*by

叉积：a^b=ax*by-bx*ay

有了这些，代码就呼之欲出了。

首先read(线段)

然后read(点)

发现满足二分性(点A一定在前t条线后，在后n-t条线前)

于是，二分优化查找

O(mlogn)

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define file(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
inline void read(int &ans){
  ans=0;char x=getchar();bool f=0;
  while(x<‘0‘||x>‘9‘){if(x==‘-‘)f=1;x=getchar();}
  while(x>=‘0‘&&x<=‘9‘)ans=ans*10+x-‘0‘,x=getchar();
  if(f)ans=-ans;
}
typedef int db;//此题不用开double
const int MAXN=5000+10;
struct Po{//定义点(向量)
  db x,y;
  Po(){};
  Po(int x,int y):x(x),y(y){}
  Po operator-(const Po A)const{return Po(x-A.x,y-A.y);}
  db operator*(const Po A)const{return x*A.y-y*A.x;}
};
int n,m,ans[MAXN],first=true;
db X1,Y1,X2,Y2,U[MAXN],L[MAXN];

bool check(Po C,int ID){
  Po A=Po(L[ID],Y2);
  Po B=Po(U[ID],Y1);
  return (C-A)*(B-A)>0;
}

int find(db x,db y){
  int l=0,r=n+1;
  while(l<r){
    int mid=(l+r)>>1;
    if(check(Po(x,y),mid))l=mid+1;
    else r=mid;
  }
  return l-1;
}

db x,y;
int main(){
  while(scanf("%d",&n)&&n){
    if(first)first=false;else if(n)puts("");
    memset(ans,0,sizeof(ans));
    read(m),read(X1),read(Y1),read(X2),read(Y2);
    U[n]=L[0]=X1,U[n+1]=L[n+1]=X2;
    for(int i=1;i<=n;i++)read(U[i]),read(L[i]);
    for(int i=0;i<m;i++){
      read(x),read(y);
      ans[find(x,y)]++;
    }
    for(int i=0;i<=n;i++)printf("%d: %d\n",i,ans[i]);
  }
  return 0;
}