hdu1542----Atlantis

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7541    Accepted Submission(s): 3318

Problem Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

 

Input

The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.

The input file is terminated by a line containing a single 0. Don’t process it.

 

Output

For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.

Output a blank line after each test case.

 

Sample Input

   
   
   
   

    
    
    
    2
10 10 20 20
15 15 25 25.5
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Test case #1
Total explored area: 180.00 
   
   
   
   

 

Source

Mid-Central European Regional Contest 2000

 

Recommend

线段树+扫描线+离散化求矩形面积并
方法就不细说了，就是离散化以后的线段右端要-1， 防止某两个矩形的边重合的情况，如果不减一，插入倒线段树上会导致那个点被覆盖的次数变多

/*************************************************************************
    > File Name: hdu1542.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年01月15日 星期四 11时13分38秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 210;

int n, cnt;
struct node
{
	double l, r;
	double h;
	int flag;
}lines[N << 1];

struct seg
{
	int l, r;
	int add;
	double len;
}tree[N << 2];

double xis[N];

int cmp (node a, node b)
{
	return a.h < b.h;
}

int BinSearch (double val)
{
	int l = 1;
	int r = cnt;
	int mid;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (xis[mid] < val)
		{
			l = mid + 1;
		}
		else if (xis[mid] > val)
		{
			r = mid - 1;
		}
		else
		{
			break;
		}
	}
	return mid;
}

void build (int p, int l, int r)
{
	tree[p].l = l;
	tree[p].r = r;
	tree[p].len = 0;
	tree[p].add = 0;
	if (l == r)
	{
		return;
	}
	int mid = (l + r) >> 1;
	build (p << 1, l, mid);
	build (p << 1 | 1, mid + 1, r);
}

void pushup (int p)
{
	if (tree[p].add)
	{
		tree[p].len = xis[tree[p].r + 1] - xis[tree[p].l]; 
	}
	else if (tree[p].l == tree[p].r)
	{
		tree[p].len = 0;
	}
	else
	{
		tree[p].len = tree[p << 1].len + tree[p << 1 | 1].len;
	}
}

void update (int p, int l, int r, int val)
{
	if (l == tree[p].l && r == tree[p].r)
	{
		tree[p].add += val;
		pushup(p);
		return;
	}
	//这里无需down，因为对应每一条直线，插入之后一定有删除操作
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (r <= mid)
	{
		update (p << 1, l, r, val);
	}
	else if (l > mid)
	{
		update (p << 1 | 1, l, r, val);
	}
	else
	{
		update (p << 1, l, mid, val);
		update (p << 1 | 1, mid + 1, r, val);
	}
	pushup (p);
}

int main()
{
	int icase = 1;
	while (~scanf("%d", &n), n)
	{
		double x1, y1, x2, y2;
		cnt = 0;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
			lines[i].flag = -1;
			lines[i].l = x1;
			lines[i].r = x2;
			lines[i].h = y2;
			xis[++cnt] = x1;
			xis[++cnt] = x2;
			lines[i + n].l = x1;
			lines[i + n].r = x2;
			lines[i + n].h = y1;
			lines[i + n].flag = 1;
		}
		sort (lines + 1, lines + 2 * n + 1, cmp);
		sort (xis + 1, xis + cnt + 1);
		cnt = unique (xis + 1, xis + 1 + cnt) - xis - 1;
		double ans = 0;
		build (1, 1, cnt);
		int l = BinSearch (lines[1].l);
		int r = BinSearch (lines[1].r) - 1;
		update (1, l, r, lines[1].flag);
		for (int i = 2; i <= 2 * n; ++i)
		{
			ans += tree[1].len * (lines[i].h - lines[i - 1].h);
			l = BinSearch (lines[i].l);
			r = BinSearch (lines[i].r) - 1;
			update (1, l, r, lines[i].flag);
		}
		printf("Test case #%d\n", icase++);
		printf("Total explored area: %.2f\n\n", ans);
	}
	return 0;
}