Atlantis
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 16222   Accepted: 6172

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 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 思路:首先将横坐标离散化,再对纵坐标排序,然后根据y轴从下往上扫描,每次的高度就是seg[i].y-seg[i-1].y,这就相当于分矩形的宽,然后要做的事就是查询x轴(矩形长)的有效长度,这就要交给线段树了。 AC代码:
 /*************************************************************************
> File Name: area.cpp
> Author: wangzhili
> Mail: wangstdio.h@gmail.com
> Created Time: 2014/3/1 星期六 16:06:57
************************************************************************/ #include<iostream>
#include<algorithm>
#include<cstdio>
#define MAX 1000
using namespace std;
class TreeNode
{
public:
int left;
int right;
int mid;
int cover;
int flag;
double length;
};
typedef struct
{
double xl, xr, y;
int flag;
}Line;
TreeNode node[*MAX];
Line seg[MAX];
double x[MAX];
double length;
bool cmp(Line a, Line b)
{
return a.y < b.y;
} void BuildTree(int k, int l, int r)
{
node[k].left = l;
node[k].right = r;
node[k].mid = (l + r) >> ;
node[k].cover = ;
node[k].flag = ;
if(l + == r)
{
node[k].flag = ;
return ;
}
int mid = (l + r) >> ;
BuildTree(k << , l, mid);
BuildTree(k << |, mid, r);
} void UpdateTree(int k, int l, int r, int flag)
{
if(node[k].left == l && node[k].right == r)
{
node[k].cover += flag;
node[k].length = x[r-] - x[l-];
return ;
}
if(node[k].flag)
return ;
if(node[k].mid <= l)
UpdateTree(k << |, l, r, flag);
else if(node[k].mid >= r)
UpdateTree(k << , l, r, flag);
else
{
UpdateTree(k << , l, node[k].mid, flag);
UpdateTree(k << |, node[k].mid, r, flag);
}
} void GetLength(int k)
{
if(node[k].cover > )
{
length += node[k].length;
return ;
}
if(node[k].flag)
return;
GetLength(k << );
GetLength(k << |);
} int GetIndex(double num, int length)
{
int l, r, mid;
l = , r = length-;
while(l <= r)
{
mid = (l + r) >> ;
if(x[mid] == num)
return mid;
else if(x[mid] > num)
r = mid - ;
else
l = mid + ;
}
} int main(int argc, char const *argv[])
{
int n, i, j, k, cnt;
int xl, xr;
double ans;
double x1, y1, x2, y2;
cnt = ;
BuildTree(, , );
// freopen("in.c", "r", stdin);
while(~scanf("%d", &n) && n)
{
j = ;
ans = .;
for(i = ; i < ; i ++)
{
node[i].cover = ;
}
for(i = ; i < n; i ++)
{
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
seg[j].xl = x1;
seg[j].xr = x2;
seg[j].y = y1;
x[j] = x1;
seg[j ++].flag = ;
seg[j].xl = x1;
seg[j].xr = x2;
seg[j].y = y2;
x[j] = x2;
seg[j ++].flag = -;
}
sort(x, x+j);
sort(seg, seg+j, cmp);
k = ;
for(i = ; i < j; i ++)
{
if(x[i] != x[i-])
x[k ++] = x[i];
}
xl = GetIndex(seg[].xl, k) + ;
xr = GetIndex(seg[].xr, k) + ;
UpdateTree(, xl, xr, seg[].flag);
length = ;
GetLength();
for(i = ; i < j; i ++)
{
ans += (seg[i].y-seg[i-].y)*length;
xl = GetIndex(seg[i].xl, k)+;
xr = GetIndex(seg[i].xr, k)+;
UpdateTree(, xl, xr, seg[i].flag);
length = .;
GetLength();
}
printf("Test case #%d\nTotal explored area: %.2lf\n\n", ++cnt, ans);
}
return ;
}
												

POJ -- 1151的更多相关文章

  1. 扫描线三巨头 hdu1928&&hdu 1255 && hdu 1542 [POJ 1151]

    学习链接:http://blog.csdn.net/lwt36/article/details/48908031 学习扫描线主要学习的是一种扫描的思想,后期可以求解很多问题. 扫描线求矩形周长并 hd ...

  2. POJ 1151 Atlantis(扫描线)

    题目原链接:http://poj.org/problem?id=1151 题目中文翻译: POJ 1151 Atlantis Time Limit: 1000MS   Memory Limit: 10 ...

  3. POJ 1151 Atlantis(线段树-扫描线,矩形面积并)

    题目链接:http://poj.org/problem?id=1151 题目大意:坐标轴上给你n个矩形, 问这n个矩形覆盖的面积 题目思路:矩形面积并. 代码如下: #include<stdio ...

  4. POJ 1151 Atlantis (扫描线+线段树)

    题目链接:http://poj.org/problem?id=1151 题意是平面上给你n个矩形,让你求矩形的面积并. 首先学一下什么是扫描线:http://www.cnblogs.com/scau2 ...

  5. POJ 1151 Atlantis 矩形面积求交/线段树扫描线

    Atlantis 题目连接 http://poj.org/problem?id=1151 Description here are several ancient Greek texts that c ...

  6. poj 1151(离散化+矩形面积并)

    题目链接:http://poj.org/problem?id=1151 关于离散化,这篇博客讲的很好:http://www.cppblog.com/MiYu/archive/2010/10/15/12 ...

  7. 【POJ 1151】Atlantis

    离散化后扫描线扫一遍. 夏令营时gty学长就讲过扫描线,可惜当时too naive,知道现在才写出模板题. 当时也不会线段树啊233 #include<cstdio> #include&l ...

  8. POJ 1151 Atlantis 线段树+离散化+扫描线

    这次是求矩形面积并 /* Problem: 1151 User: 96655 Memory: 716K Time: 0MS Language: G++ Result: Accepted */ #inc ...

  9. [POJ 1151] Atlantis

    一样的题:HDU 1542 Atlantis Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18148   Accepted ...

  10. hdu 1542&&poj 1151 Atlantis[线段树+扫描线求矩形面积的并]

    Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total S ...

随机推荐

  1. 浅谈break 、continue、return,goto四种语句的区别。

    浅谈break .continue.return三种语句的区别: break,continue,return这三个具有跳转功能的语句在c语言中经常被用到,近期身边有些小伙伴总是把它们的用法搞乱,在这里 ...

  2. ZOJ 1013 Great Equipment(DP)

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=13 题目大意:说的是有三种不同的装备,分别是头盔,盔甲,战靴需要运输, ...

  3. 一段sql的优化

    优化前代码 select * ,ROW_NUMBER() OVER(order by WrongCount desc) as rowId from(select Quba_IDint,Quba_Num ...

  4. ubuntu ssh安装

    参考 http://www.linuxidc.com/Linux/2010-02/24349.htm  文章很不错!! ssh   登录名@ip地址 , 如果提示验证key can't be esta ...

  5. SVN资料库转移-----dump和load

    最近由于大批量的更换服务器,所以之前布署的SVN服务器需要重新布署,需要把原来的资源库转移到新服务器上,并且使管理的项目版本一致,在网上查了一下SVN版本库迁移,但看了一上google出来的也很少,所 ...

  6. Javascript数组的indexOf()、lastIndexOf()方法

    在javascript数组中提供了两个方法来对数组进行查找,这两个方法分别为indexOf(),lastIndexOf(). 这两个方法都有两个参数,第一个参数为需要查找的项,第二个参数则是查找的起始 ...

  7. mac 生成支付宝的rsa公钥和私钥 php版本

    openssl genrsa -out rsa_private_key.pem 1024 公钥 openssl rsa -in rsa_private_key.pem -pubout -out rsa ...

  8. WPF的依赖属性

    Windows Presentation Foundation (WPF) 提供了一组服务,这些服务可用于扩展公共语言运行时 (CLR)属性的功能,这些服务通常统称为 WPF 属性系统.由 WPF 属 ...

  9. 转:浅谈命令查询职责分离(CQRS)模式

    原文来自于:http://www.cnblogs.com/yangecnu/p/Introduction-CQRS.html 在常用的三层架构中,通常都是通过数据访问层来修改或者查询数据,一般修改和查 ...

  10. IntelliJ 一键添加双引号

    IntelliJ IDEA(目前最新版本为2016.2.5) 不直接支持一键向选取内容添加双引号,但可以通过配置 Live Template 实现此功能. 以下为配置方法(针对2016.2.5版,其它 ...