Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10208    Accepted Submission(s): 4351

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
/*
hdu 1542 线段树扫描(面积) 给你n个矩形,求最终形成的图形的面积大小 数据不一定是整数,所以先对他们进行离散化处理
大致就是每次计算平行于x轴的两条相邻线之间的面积,我们已经知道了两条平行
线之间的高度,于是就转变成在求当前情况下映射到x轴上的线段的长度,这个便能
利用线段树解决了 先把所有平行于x轴的线段按高度排序,然后从下往上,每次在遇到矩形下边时在
[l,r]上加1表示线段覆盖,遇到上边则填上-1消除影响.然后每次计算覆盖长度
再乘上高即可 参考(图文详解):
http://www.cnblogs.com/scau20110726/archive/2013/04/12/3016765.html hhh-2016-03-26 17:58:50
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
const int maxn = 10005;
double hs[maxn];
struct node
{
int l,r;
double len;
int sum;
int mid()
{
return (l+r)>>1;
}
} tree[maxn*5]; void push_up(int i)
{
if(tree[i].sum)
{
tree[i].len = (hs[tree[i].r+1]-hs[tree[i].l]);
}
else if(tree[i].l == tree[i].r)
{
tree[i].len= 0;
}
else
{
tree[i].len = tree[lson].len+tree[rson].len;
}
} void build(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].sum = tree[i].len = 0;
if(l == r)
return ; int mid = tree[i].mid();
build(lson,l,mid);
build(rson,mid+1,r);
push_up(i);
} void push_down(int i)
{ } void Insert(int i,int l,int r,int val)
{
if(tree[i].l >= l && tree[i].r <=r )
{
tree[i].sum += val;
push_up(i);
return ;
}
int mid = tree[i].mid();
push_down(i);
if(l <= mid)
Insert(lson,l,r,val);
if(r > mid)
Insert(rson,l,r,val);
push_up(i);
} struct edge
{
double l,r,high;
int va;
edge() {};
edge(double _l,double _r,double _high,int _va):l(_l),r(_r),high(_high),va(_va)
{}
};
edge tx[maxn*2]; bool cmp(edge a,edge b)
{
if(a.high != b.high)
return a.high < b.high;
else
return a.va > b.va;
}
int tot,m;
int fin(double x)
{
int l = 0,r = m-1;
while(l <= r)
{
int mid = (l+r)>>1;
if(hs[mid] == x)
return mid;
else if(hs[mid] < x)
l = mid+1;
else
r = mid-1;
}
} int main()
{
int n;
int cas =1;
while(scanf("%d",&n) != EOF && n)
{
double x1,x2,y1,y2;
tot = 0;
for(int i = 1; i <= n; i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
hs[tot] = x1;
tx[tot++] = edge(x1,x2,y1,1);
hs[tot] = x2;
tx[tot++] = edge(x1,x2,y2,-1);
}
sort(tx,tx+tot,cmp);
sort(hs,hs+tot);
m = 1;
for(int i = 1;i < tot;i++)
{
if(hs[i] != hs[i-1])
hs[m++] = hs[i];
}
build(1,0,m);
double ans = 0;
for(int i = 0;i < tot;i++)
{
int l = fin(tx[i].l);
int r = fin(tx[i].r)-1; Insert(1,l,r,tx[i].va);
ans += (tree[1].len)*(tx[i+1].high-tx[i].high);
//cout << ans <<endl;
}
printf("Test case #%d\n",cas++);
printf("Total explored area: %.2f\n\n",ans);
}
return 0;
}

  

												

hdu 1542 线段树扫描(面积)的更多相关文章

  1. hdu 1542 线段树+扫描线 学习

    学习扫描线ing... 玄学的东西... 扫描线其实就是用一条假想的线去扫描一堆矩形,借以求出他们的面积或周长(这一篇是面积,下一篇是周长) 扫描线求面积的主要思想就是对一个二维的矩形的某一维上建立一 ...

  2. hdu 1542 线段树之扫描线之面积并

    点击打开链接 题意:给你n个矩形,求它们的面积,反复的不反复计算 思路:用线段树的扫描线完毕.将X坐标离散化后,从下到上扫描矩形,进行各种处理,看代码凝视把 #include <stdio.h& ...

  3. Atlantis HDU - 1542 线段树+扫描线 求交叉图形面积

    //永远只考虑根节点的信息,说明在query时不会调用pushdown //所有操作均是成对出现,且先加后减 // #include <cstdio> #include <cstri ...

  4. HDU 1542 线段树离散化+扫描线 平面面积计算

    也是很久之前的题目,一直没做 做完之后觉得基本的离散化和扫描线还是不难的,由于本题要离散x点的坐标,最后要计算被覆盖的x轴上的长度,所以不能用普通的建树法,建树建到r-l==1的时候就停止,表示某段而 ...

  5. HDU 1542 线段树+扫描线+离散化

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

  6. hdu 1542 线段树 求矩形并

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

  7. hdu 1828 线段树扫描线(周长)

    Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Sub ...

  8. hdu 4052 线段树扫描线、奇特处理

    Adding New Machine Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Othe ...

  9. hdu 4533 线段树(问题转化+)

    威威猫系列故事——晒被子 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Tot ...

随机推荐

  1. 201621123043《java程序设计》第4周学习总结

    1. 本周学习总结 1.1 写出你认为本周学习中比较重要的知识点关键词 关键字:继承.覆盖.多态 1.2 尝试使用思维导图将这些关键词组织起来.注:思维导图一般不需要出现过多的字. 1.3 可选:使用 ...

  2. Android Studio使用过程中遇到的错误

    > 错误1 1. This fragment should provide a default constructor (a public constructor wit 代码不规范,这个错误是 ...

  3. UWP 页面间传递参数(常见类型string、int以及自定义类型)

    这是一篇很基础的,大佬就不要看了,也不要喷,谢谢

  4. nyoj 阶乘0

    阶乘的0 时间限制:3000 ms  |  内存限制:65535 KB 难度:3   描述 计算n!的十进制表示最后有多少个0   输入 第一行输入一个整数N表示测试数据的组数(1<=N< ...

  5. print 函数设置字体颜色

    格式:\033[显示方式;前景色;背景色m数值表示的参数含义:显示方式: 0(默认值).1(高亮).22(非粗体).4(下划线).24(非下划线). 5(闪烁).25(非闪烁).7(反显).27(非反 ...

  6. ssh整合之二hibernate单独搭建

    1.首先我们需要去拷贝我们的hibernate所需的jar包  这里还需要加入我们C3P0的jar包,因为我们hibernate中使用的C3P0连接池 2. 编写我们的关系映射文件Customer.c ...

  7. JsonCPP库使用

    1.使用环境DevC++ a.建立C++工程,并添加.\JsonCPP\jsoncpp-master\jsoncpp-master\src\lib_json中源文件到工程中. b.添加头文件路径 2. ...

  8. 日推20单词 Day01

    1.conflict n. 冲突 2.electronic adj. 电子的 3.mine n. 矿藏,地雷 4.mineral n. 矿物质 adj. 矿物的 5.undermine vt 破坏,渐 ...

  9. zipline-benchmarks.py文件改写

    改写原因:在这个模块中的 get_benchmark_returns() 方法回去谷歌财经下载对应SPY(类似于上证指数)的数据,但是Google上下载的数据在最后写入Io操作的时候会报一个恶心的编码 ...

  10. pandas笔记

    axis = 1表示按列的方向遍历 axis = 0表示按行的方向遍历 Usually axis=0 is said to be "column-wise" (and axis=1 ...