（计算几何）POJ2318TOYS、POJ2398Toy Storage

POJ2318TOYS

题意：

给一个矩形和$n$条线将矩形分成$n+1$块区域，给$m$个点保证点不在线上，问每一块区域里有几个点。

思路：

很容易看出，一个点如果在一个区域内，它会在一条边的顺时针方向和另一条边的逆时针方向。所以用叉积。
如果一个点在一个区域里，设区域为$ABCD$，点为$P$。可得$(\overrightarrow{AB}\times \overrightarrow{AP})\ast (\overrightarrow{CD}\times \overrightarrow{CP})<0$。暴力枚举即可。

代码：

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<ctype.h>
#include<queue>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#define pii pair<int,int>
#define ll long long
#define cl(x,y) memset(x,y,sizeof(x))
#define ct cerr<<"Time elapsed:"<<1.0*clock()/CLOCKS_PER_SEC<<"s.\n";
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define lson x<<1,l,mid
#define rson x<<1|1,mid+1,r
#define INF 1e18
const int N=1e6+10;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1);
using namespace std;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0) return -1;
    else return 1;
}
struct point
{
    double x,y;
    point(){}
    point(double xx,double yy){x=xx;y=yy;}
    point operator +(point b){return point(x+b.x,y+b.y);}
    point operator -(point b){return point(x-b.x,y-b.y);}
    double operator ^(point b){return x*b.y-y*b.x;}//叉乘 
    double operator *(point b){return x*b.x+y*b.y;}//点乘 
    point operator *(double b){return point(x*b,y*b);}//数乘   
    void read(){scanf("%lf%lf",&x,&y);}
}p;
struct line
{
    point s,e,l;
    line(){}
    line(point ss,point ee){s = ss;e = ee;l=e-s;}
    void read(){s.read(),e.read();l=e-s;}
}l[5010];
int ans[5010]={0};
int main()
{
	int n,m;
	double x1,y1,x2,y2;
	while(~scanf("%d",&n) && n)
	{
		int i,j;
		cl(ans,0);
		scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
		l[0]={point{x1,y1},point{x1,y2}};
		l[n+1]={point{x2,y1},point{x2,y2}};
		for(i=1;i<=n;i++)
		{
			double u,v;
			scanf("%lf%lf",&u,&v);
			l[i]={point{u,y1},point{v,y2}};
		}
		for(i=1;i<=m;i++)
		{
			p.read();
			for(j=0;j<=n;j++)
				if(((p-l[j].s)^l[j].l)*((p-l[j+1].s)^l[j+1].l)<0)
				{
					ans[j]++;
					break;
				}				
		}
		for(i=0;i<=n;i++)
			printf("%d: %d\n",i,ans[i]);
		puts("");
	}
	return 0;
}

POJ2398Toy Storage

题意：

与上一题基本一致，就是输出不同，从小到大输出含有点的个数的格子有多少个。

思路：

与上一题基本一致，就是注意给的线要排序。

代码：

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<ctype.h>
#include<queue>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#define pii pair<int,int>
#define ll long long
#define cl(x,y) memset(x,y,sizeof(x))
#define ct cerr<<"Time elapsed:"<<1.0*clock()/CLOCKS_PER_SEC<<"s.\n";
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define lson x<<1,l,mid
#define rson x<<1|1,mid+1,r
#define INF 1e18
const int N=1e6+10;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1);
using namespace std;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0) return -1;
    else return 1;
}
struct point
{
    double x,y;
    point(){}
    point(double xx,double yy){x=xx;y=yy;}
    point operator +(point b){return point(x+b.x,y+b.y);}
    point operator -(point b){return point(x-b.x,y-b.y);}
    double operator ^(point b){return x*b.y-y*b.x;}//叉乘 
    double operator *(point b){return x*b.x+y*b.y;}//点乘 
    point operator *(double b){return point(x*b,y*b);}//数乘   
    void read(){scanf("%lf%lf",&x,&y);}
}p;
struct line
{
    point s,e,l;
    line(){}
    line(point ss,point ee){s = ss;e = ee;l=e-s;}
    void read(){s.read(),e.read();l=e-s;}
}l[5010];
int ans[5010]={0},sum[5010];
int cmp(line a,line b)
{
	if(a.s.x==b.s.x)
		return a.e.x<b.e.x;
	return a.s.x<b.s.x;
}
int main()
{
	int n,m;
	double x1,y1,x2,y2;
	while(~scanf("%d",&n) && n)
	{
		int i,j;
		cl(ans,0);
		cl(sum,0);
		scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
		l[0]={point{x1,y1},point{x1,y2}};
		l[n+1]={point{x2,y1},point{x2,y2}};
		for(i=1;i<=n;i++)
		{
			double u,v;
			scanf("%lf%lf",&u,&v);
			l[i]={point{u,y1},point{v,y2}};
		}
		sort(l,l+n+2,cmp);
		for(i=1;i<=m;i++)
		{
			p.read();
			for(j=0;j<=n;j++)
				if(sgn((p-l[j].s)^l[j].l)*sgn((p-l[j+1].s)^l[j+1].l)<0)
				{
					ans[j]++;
					break;
				}				
		}
		for(i=0;i<=n;i++)
			sum[ans[i]]++;
		printf("Box\n");
		for(i=1;i<=m;i++)
			if(sum[i])
				printf("%d: %d\n",i,sum[i]);
	}
	return 0;
}