Shaping Regions

Time limit: 0.5 second
Memory limit: 64 MB
N opaque rectangles (1 ≤ N ≤ 1000) of various colors are placed on a white sheet of paper whose size is A wide by B long. The rectangles are put with their sides parallel to the sheet's borders. All rectangles fall within the borders of the sheet so that different figures of different colors will be seen.
The coordinate system has its origin (0, 0) at the sheet's lower left corner with axes parallel to the sheet's borders.

Input

The order of the input lines dictates the order of laying down the rectangles. The first input line is a rectangle “on the bottom”. First line contains AB and N, space separated (1 ≤ AB ≤ 10000). Lines 2, …, N + 1 contain five integers each: llxllyurxury, color: the lower left coordinates and upper right coordinates of the rectangle whose color is color (1 ≤ color ≤ 2500) to be placed on the white sheet. The color 1 is the same color of white as the sheet upon which the rectangles are placed.

Output

The output should contain a list of all the colors that can be seen along with the total area of each color that can be seen (even if the regions of color are disjoint), ordered by increasing color. Do not display colors with no area.

Sample

input output
20 20 3
2 2 18 18 2
0 8 19 19 3
8 0 10 19 4
1 91
2 84
3 187
4 38

分析:经典的覆盖问题,参考http://blog.csdn.net/skyprophet/article/details/4514926,冰块上浮法;

   倒序计算,在计算到当前矩形时,看在他上面的矩形有没有重叠的,有就去掉那个部分,一直递归下去即可;

代码:  

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=3e3+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,k,t,ans[maxn];
struct node
{
int x1,x2,y1,y2,c;
node(){}
node(int _x1,int _x2,int _y1,int _y2,int _c)
{
x1=_x1,x2=_x2,y1=_y1,y2=_y2,c=_c;
}
}op[maxn];
int get(node p,int nt)
{
int ans=;
while(nt<=k&&((p.x1>=op[nt].x2)||(p.x2<=op[nt].x1)||(p.y1>=op[nt].y2)||(p.y2<=op[nt].y1)))
nt++;
if(nt>k)return (p.x2-p.x1)*(p.y2-p.y1);
if(p.x1<op[nt].x1)
ans+=get(node(p.x1,op[nt].x1,p.y1,p.y2,op[nt].c),nt+),p.x1=op[nt].x1;
if(p.x2>op[nt].x2)
ans+=get(node(op[nt].x2,p.x2,p.y1,p.y2,op[nt].c),nt+),p.x2=op[nt].x2;
if(p.y1<op[nt].y1)
ans+=get(node(p.x1,p.x2,p.y1,op[nt].y1,op[nt].c),nt+),p.y1=op[nt].y1;
if(p.y2>op[nt].y2)
ans+=get(node(p.x1,p.x2,op[nt].y2,p.y2,op[nt].c),nt+),p.y2=op[nt].y2;
return ans;
}
int main()
{
int i,j;
scanf("%d%d%d",&n,&m,&k);
ans[]=n*m;
rep(i,,k)scanf("%d%d%d%d%d",&op[i].x1,&op[i].y1,&op[i].x2,&op[i].y2,&op[i].c);
for(i=k;i>=;i--)
{
ans[op[i].c]+=(j=get(op[i],i+));
ans[]-=j;
}
rep(i,,)if(ans[i])printf("%d %d\n",i,ans[i]);
//system("Pause");
return ;
}

ural1147 Shaping Regions的更多相关文章

  1. ural 1147. Shaping Regions

    1147. Shaping Regions Time limit: 0.5 secondMemory limit: 64 MB N opaque rectangles (1 ≤ N ≤ 1000) o ...

  2. Shaping Regions(dfs)

    Shaping Regions Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 124  Solved: 39[Submit][Status][Web B ...

  3. USACO 6.2 Shaping Regions

    Shaping Regions N opaque rectangles (1 <= N <= 1000) of various colors are placed on a white s ...

  4. OI暑假集训游记

    莞中OI集训游记 Written BY Jum Leon. I        又是一载夏,本蒟蒻以特长生考入莞中,怀着忐忑的心情到了8月,是集训之际.怀着对算法学习的向往心情被大佬暴虐的一丝恐惧来到了 ...

  5. USACO 完结的一些感想

    其实日期没有那么近啦……只是我偶尔还点进去造成的,导致我没有每一章刷完的纪念日了 但是全刷完是今天啦 讲真,题很锻炼思维能力,USACO保持着一贯猎奇的题目描述,以及尽量不用高级算法就完成的题解……例 ...

  6. [LeetCode] Surrounded Regions 包围区域

    Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured ...

  7. 验证LeetCode Surrounded Regions 包围区域的DFS方法

    在LeetCode中的Surrounded Regions 包围区域这道题中,我们发现用DFS方法中的最后一个条件必须是j > 1,如下面的红色字体所示,如果写成j > 0的话无法通过OJ ...

  8. Leetcode: Surrounded regions

    Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'. A region is captured ...

  9. LEETCODE —— Surrounded Regions

    Total Accepted: 43584 Total Submissions: 284350 Difficulty: Medium Given a 2D board containing 'X' a ...

随机推荐

  1. Load PE from memory(反取证)(未完)

      Article 1:Loading Win32/64 DLLs "manually" without LoadLibrary() The most important step ...

  2. document.domain的修改问题

    有时候,需要修改document.domain. 典型的情形:http://a.xxx.com/A.htm 的主页面有一个<iframe src="http://b.xxx.com/B ...

  3. UIView animateWithDuration 使用详解

    在ios4.0及以后鼓励使用animateWithDuration方法来实现动画效果.当然,以往的begin/commit的方法依然使用,下面详细解释一下animateWithDuration的使用方 ...

  4. jquery checkbox 操作

    1.jquery 获取所有选中和未选中的checkbox 未选中 var unCheckedBoxs = $("input[name='myCheckbox']").not(&qu ...

  5. .htaccess文件url重写小记

    .htaccess文件url重写 当上一条规则匹配 并转换后 符合下一条规则的 继续下一条的匹配转换 RewriteRule ^shangpin-([0-9a-zA-Z]+)/category-([0 ...

  6. 删除sql计划 调用的目标发生了异常。 (mscorlib) 其他信息: 用户 'sa' 登录失败。

    在删除以前创建的sql的计划任务时,弹出如题错误提示,发现错误原因在于,sa密码更改过,导致在删除时因为sa的密码和当前的密码不正确出现此错误. 解决办法: 1.在计划任务的编辑窗口,找到管理连接 2 ...

  7. Inno Setup入门(二十)——Inno Setup类参考(6)

    存储框 存储框也是典型的窗口可视化组件,同编辑框类似,可以输入.显示文本,但是和编辑框不同的是,编辑框只能编辑.显示单行文本,而存储框则可以对多行文本进行操作.存储框的类定义如下:< xmlna ...

  8. launchMode传递参数注意startActivityForResult

    Activity1 到Activity2 用startActivityForResult 如果Activity2的launchMode为 singleInstance 和 singleTask 都会启 ...

  9. XHTML 与 HTML 之间的差异

    最主要的不同: XHTML 元素必须被正确地嵌套. XHTML 元素必须被关闭. 标签名必须用小写字母. XHTML 文档必须拥有根元素.

  10. openCV(二)---iOS中使用openCV的图片格式转换

    可以实现将UIImage和IplImage类型实现相互转换 //由于OpenCV主要针对的是计算机视觉方面的处理,因此在函数库中,最重要的结构体是IplImage结构. - (IplImage *)C ...