题目链接:http://poj.org/problem?id=2318
题目大意:给出一个盒子,然后从左到右给出n条线,将盒子划分成了n+1个格子,然后再给出m个点,判断每个点在那个格子中。
思路:从左上角到每一个点拉一根线,然后判断与每个格子的边缘的线是否相交。一开始我用的暴力,一个个线判断TLE,发现线是有顺序的,所以用二分,二分出第一个没有相交的线即可。
二分总结:https://blog.csdn.net/qq_40482358/article/details/84555743
ACCode:
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<string>
#include<fstream>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define Pair pair<int,int>
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// ??
//std::ios::sync_with_stdio(false);
// register
const int MAXN=5e3+10;
const int INF32=0x3f3f3f3f;
const ll INF64=0x3f3f3f3f3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);
const double EPS=1.0e-8;
struct Point{
double x,y;
Point(double _x=0,double _y=0){
x=_x;y=_y;
}
friend Point operator - (const Point &a,const Point &b){
return Point(a.x-b.x,a.y-b.y);
}
};
struct V{
Point start,end;
V(Point _start=Point(0,0),Point _end=Point(0,0)){
start=_start;end=_end;
}
};
V Line[MAXN];
Point LU,RD,Dot;
int Sum[MAXN];
int n,m;
double CroMul(V a,V b){
a.end=a.end-a.start;b.end=b.end-b.start;
return a.end.x*b.end.y-b.end.x*a.end.y;
}
int LineInter(V l1,V l2){
if(max(l1.start.x,l1.end.x)>=min(l2.start.x,l2.end.x)&&
max(l2.start.x,l2.end.x)>=min(l1.start.x,l1.end.x)&&
max(l1.start.y,l1.end.y)>=min(l2.start.y,l2.end.y)&&
max(l2.start.y,l2.end.y)>=min(l1.start.y,l1.end.y)){
if(CroMul(l2,V(l2.start,l1.start))*CroMul(l2,V(l2.start,l1.end))<=0&&
CroMul(l1,V(l1.start,l2.start))*CroMul(l1,V(l1.start,l2.end))<=0){
return 1;
}
}return 0;
}
int Judge(Point p){//判断p和多少条线相交。
int l=0,r=n-1,mid;
while(l<=r){
mid=(l+r)>>1;
if(LineInter(V(LU,p),Line[mid])==0) r=mid-1;//不相交
else l=mid+1;
}return l;
}
int main(){
while(~scanf("%d",&n)){
if(n==0) break;
scanf("%d",&m);
clean(Sum,0);
scanf("%lf%lf%lf%lf",&LU.x,&LU.y,&RD.x,&RD.y);
double U,D;
for(int i=0;i<n;++i){
scanf("%lf%lf",&U,&D);
Line[i]=V(Point(U,LU.y),Point(D,RD.y));
}
for(int i=0;i<m;++i){
scanf("%lf%lf",&Dot.x,&Dot.y);
Sum[Judge(Dot)]++;
}
for(int i=0;i<=n;++i){
printf("%d: %d\n",i,Sum[i]);
}printf("\n");
}
}
/*
Sample Input
1 1
3 8
9 31
Sample Output
0
5
11
*/