POJ 2318/2398 叉积性质

2318

2398

题意：给出n条线将一块区域分成n+1块空间，再给出m个点，询问这些点在哪个空间里。

思路：由于只要求相对位置关系，而对具体位置不关心，那么易使用叉积性质得到相对位置关系（左侧/右侧），再因为是简单几何线段不相较，即有序分布，那么在求在哪个区间时可以先对所有线段根据x坐标排序，使用二分减少复杂度。

/** @Date    : 2017-07-11 11:05:59
  * @FileName: POJ 2318 叉积性质.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
//#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

struct Point
{
	int x, y;
	Point(){}
	Point(int xx, int yy){x = xx, y = yy;}
	Point operator -(const Point &b) const
	{
		return Point(x - b.x, y - b.y);
	}
	int operator *(const Point &b) const
	{
		return x * b.x + y * b.y;
	}
};

int cross(Point a, Point b)
{
	return a.x * b.y - a.y * b.x;
}

struct Line
{
	Point s, t;
	Line(){}
	Line(Point ss, Point tt){s = ss, t = tt;}
};

int JudegeCross(Point p0, Point p1, Point p2)
{
	return cross(p1 - p0, p2 - p0);
}

Line li[N];
int ans[N];
int vis[N];
int cmp(Line a, Line b)
{
	return a.s.x < b.s.x;
}

int main()
{
	int n, m, x1, x2, y1, y2;
	while(~scanf("%d", &n) && n)
	{
		MMF(ans);
		MMF(vis);
		scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
		for(int i = 0; i < n; i++)
		{
			int s, t;
			scanf("%d%d", &s, &t);
			li[i] = Line(Point(s, y1), Point(t, y2));
		}
		li[n] = Line(Point(x2, y1), Point(x2, y2));
		sort(li, li + n + 1, cmp);
		while(m--)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			Point p = Point(x, y);
			int l = 0, r = n;
			int pos = 0;
			while(l <= r)
			{
				int mid = (l + r) >> 1;
				if(JudegeCross(p, li[mid].s, li[mid].t) < 0)
				{
					pos = mid;
					r = mid - 1;
				}
				else
					l = mid + 1;
			}
			ans[pos]++;
		}
		printf("Box\n");
		for(int i = 0; i <= n; i++)
			if(ans[i])
				vis[ans[i]]++;
		for(int i = 1; i <= n; i++)
			if(vis[i])
				printf("%d: %d\n", i, vis[i]);
	}
    return 0;
}

/** @Date    : 2017-07-11 11:05:59
  * @FileName: POJ 2318 叉积性质.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
//#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

struct Point
{
	int x, y;
	Point(){}
	Point(int xx, int yy){x = xx, y = yy;}
	Point operator -(const Point &b) const
	{
		return Point(x - b.x, y - b.y);
	}
	int operator *(const Point &b) const
	{
		return x * b.x + y * b.y;
	}
};

int cross(Point a, Point b)
{
	return a.x * b.y - a.y * b.x;
}

struct Line
{
	Point s, t;
	Line(){}
	Line(Point ss, Point tt){s = ss, t = tt;}
};

int JudegeCross(Point p0, Point p1, Point p2)
{
	return cross(p1 - p0, p2 - p0);
}

Line li[N];
int ans[N];

int main()
{
	int n, m, x1, x2, y1, y2;
	while(~scanf("%d", &n) && n)
	{
		MMF(ans);
		scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
		for(int i = 0; i < n; i++)
		{
			int s, t;
			scanf("%d%d", &s, &t);
			li[i] = Line(Point(s, y1), Point(t, y2));
		}
		li[n] = Line(Point(x2, y1), Point(x2, y2));
		while(m--)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			Point p = Point(x, y);
			int l = 0, r = n;
			int pos = 0;
			while(l <= r)
			{
				int mid = (l + r) >> 1;
				if(JudegeCross(p, li[mid].s, li[mid].t) < 0)
				{
					pos = mid;
					r = mid - 1;
				}
				else
					l = mid + 1;
			}
			ans[pos]++;
		}
		for(int i = 0; i <= n; i++)
		{
			printf("%d: %d\n", i, ans[i]);
		}
		printf("\n");
	}
    return 0;
}