HDU 3265 Posters pid=3265" target="_blank" style="">题目链接 题意:给定一些矩形海报.中间有孔.求贴海报的之后的海报覆盖面积并 思路:海报一张能够分割成4个矩形.然后就是普通的矩形面积并了,利用线段树维护就可以 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3265 给你n个中间被挖空了一个矩形的中空矩形,让你求他们的面积并. 其实一个中空矩形可以分成4个小的矩形,然后就是面积并,特别注意的是x1 == x3 || x2 == x4的时候,要特判一下,否则会RE. #include <iostream> #include <cstring> #include <cstdio> #include <algorithm>…

Problem Description Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters. However, Ted is such a picky guy th…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3265 题意:给你一张挖了洞的墙纸贴在墙上,问你总面积有多少. 挖了洞后其实就是多了几个矩形墙纸,一张墙纸挖了洞后可以分成4个小矩形至于怎么分看个人喜好,然后再求个矩形的面积并 要注意的是x1==x3||x2==x4||y1==y3||y2==y4时不用分矩形这是个小小的优化. #include <iostream> #include <cstring> #include <alg…

第一次做扫描线,然后使我对线段树的理解发生了动摇= =..这个pushup写的有点神奇.代码如下: #include <stdio.h> #include <algorithm> #include <string.h> #define t_mid (l+r>>1) #define ls (o<<1) #define rs (o<<1|1) #define lson ls,l,t_mid #define rson rs,t_mid+1,…

矩形面积并,一个拆成四个 #include<cstdio> #include<cstring> #include<cmath> #include<map> #include<algorithm> using namespace std; +; struct Seg { long long x; long long Y1,Y2;//离散化之后的坐标 long long flag; }s[*maxn]; *maxn],X2[*maxn],Y1[*ma…

Description Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters. However, Ted is such a picky guy that in ev…

Posters Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5230    Accepted Submission(s): 1220 Problem Description Ted has a new house with a huge window. In this big summer, Ted decides to decora…

http://acm.hust.edu.cn/vjudge/problem/10769 给n张海报,在每张海报上剪掉一个矩形,求面积并 把剪块的海报分成四个矩形,就是普通的求面积并问题了 #include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <a…

先和第二类做一个对比 第一类Stirling数是有正负的,其绝对值是包含n个元素的集合分作k个环排列的方法数目.递推公式为, S(n,0) = 0, S(1,1) = 1. S(n+1,k) = S(n,k-1) + nS(n,k). 边界条件: S(0 , 0) = 1 S(p , 0) = 0 p>=1 S(p , p) =1 p>=0 一些性质: S(p ,1) = 1 p>=1 S(p, 2) = 2^(p-1)– 1 p>=2 第二类Stirling数是把包含n个元素的集…