poj2318 poj2398

http://poj.org/problem?id=2318

题意不难理解,如图,我的代码是以线段作为最基本单位。其实一开始时写的代码是有一个四边形的概念的,如果点在四边形内,点和左边界的点乘积再乘以点和右边界的点乘积是负的,如图:

poj2318 poj2398_第1张图片

但是这样太慢,而且,原题目给出的是排好序的格子序列,如图:

poj2318 poj2398_第2张图片

所以可以二分查找,故有如下代码:

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

struct Point
{
	int x, y;
};

struct Line
{
	Point a, b;
} line[5005];

int cnt[5005];

int 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 BinarySearch(Point a, int n)
{
	int l, r, mid;

	l = 0; r = n - 1;

	while (l < r)
	{
		mid = (l + r) >> 1;
		if (Multi(a, line[mid].a, line[mid].b) > 0)
		{
			l = mid + 1;
		}
		else
		{
			r = mid;
		}
	}
	if (Multi(a, line[l].a, line[l].b) < 0)
	{
		cnt[l]++;
	}
	else
	{
		cnt[l + 1]++;
	}
}

int main()
{
	int n, m, x1, y1, x2, y2;
	int i, t1, t2;
	Point a;

	while (cin >> n && n)
	{
		cin >> m >> x1 >> y1 >> x2 >> y2;
		for (i = 0; i < n; i++)
		{
			cin >> t1 >> t2;
			line[i].a.x = t1;
			line[i].a.y = y1;
			line[i].b.x = t2;
			line[i].b.y = y2;
		}

		memset(cnt, 0, sizeof (cnt));

		for (i = 0; i < m; i++)
		{
			cin >> a.x >> a.y;
			BinarySearch(a, n);
		}

		for (i = 0; i <= n; i++)
		{
			printf("%d: %d\n", i, cnt[i]);
		}
		printf("\n");
	}
}

http://poj.org/problem?id=2398

这个是打乱顺序的,输出的时候加个统计功能就好。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

struct Point
{
	long long int x, y;
};

struct Line
{
	Point a, b;
} line[5005];

long long int cnt[1005], cnt1[1010];

int Multi(Point p1, Point p2, Point p0)
{
	return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}

bool comp (Line a, Line b)
{
	return a.a.x < b.a.x;
}

void BinarySearch(Point a, int n)
{
	int l, r, mid;

	l = 0; r = n - 1;

	while (l < r)
	{
		mid = (l + r) >> 1;
		if (Multi(a, line[mid].a, line[mid].b) > 0)
		{
			l = mid + 1;
		}
		else
		{
			r = mid;
		}
	}
	if (Multi(a, line[l].a, line[l].b) < 0)
	{
		cnt[l]++;
	}
	else
	{
		cnt[l + 1]++;
	}
}

int main()
{
	int n, m, x1, y1, x2, y2;
	int i, t1, t2;
	Point a;

	while (cin >> n && n)
	{
		cin >> m >> x1 >> y1 >> x2 >> y2;
		for (i = 0; i < n; i++)
		{
			cin >> t1 >> t2;
			line[i].a.x = t1;
			line[i].a.y = y1;
			line[i].b.x = t2;
			line[i].b.y = y2;
		}

		sort(line, line + n, comp);

		memset(cnt, 0, sizeof (cnt));
		memset(cnt1, 0, sizeof(cnt1));
		for (i = 0; i < m; i++)
		{
			cin >> a.x >> a.y;
			BinarySearch(a, n);
		}

		cout << "Box" << endl;
		for (int i = 0; i <= n; i++)
		{
			if (cnt[i])
			{
				cnt1[cnt[i]]++;
			}
		}
		for (int i = 1; i <= m; i++)
		{
			if (cnt1[i])
			{
				printf("%d: %I64d\n", i, cnt1[i]);
			}
		}
	}
}


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