poj2398 Toy Storage 链接 poj 题目大意 这道题的大概意思是先输入6个数字:n,m,x1,y1,x2,y2.n代表卡片的数量,卡片竖直(或倾斜)放置在盒内,可把盒子分为n+1块区域,然后分别用0到n表示,m代表玩具的个数,(x1,y1)代表盒子的左上顶点坐标,(x2,y2)代表盒子的右下顶点坐标,接下来的n行,每行都有两个数a,b,(a,y1),(b,y2)分别代表卡片的两个顶点位置,接下来的m行每行两个数从c,d,(c,d),代表玩具放置的坐标,最后让你输出当区域内玩具的…

Toy Storage Time Limit: 1000MS Memory Limit: 65536K Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is…

题目链接 Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4104   Accepted: 2433 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangula…

题目传送门:poj 2398 Toy Storage 题目大意:一个长方形的箱子,里面有一些隔板,每一个隔板都可以纵切这个箱子.隔板将这个箱子分成了一些隔间.向其中扔一些玩具,每个玩具有一个坐标,求有\(t​\)个玩具的隔间数(对\(t>0​\)都要输出). 题目分析:涉及到计算几何的知识是求点在线的哪一侧.可以利用叉积来做.取点\(A\)到隔板的上端点\(B\)的向量\(\vec{AB}\)叉乘点\(A\)到隔板的下端点\(C\)的向量\(\vec{AC}\).叉积的公式\(\vec a\ti…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4588   Accepted: 2718 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5016   Accepted: 2978 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

题目链接:http://poj.org/problem?id=2398 思路RT,和POJ2318一样,就是需要排序,输出也不一样.手工画一下就明白了.注意叉乘的时候a×b是判断a在b的顺时针还是逆时针侧,>0是顺时针测,<0是逆时针侧,本题对应看成右.左侧,特别注意. /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリキリ♂ mind! ┛┗┛┗┛┃＼○/ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓…

题意:给定一个如上的长方形箱子,中间有n条线段,将其分为n+1个区域,给定m个玩具的坐标,统计每个区域中的玩具个数. 题解:通过斜率判断一个点是否在两条线段之间. /** 通过斜率比较点是否在两线段之间 */ #include"iostream" #include"cstdio" #include"algorithm" #include"cstring" using namespace std; const int N=100…

二分点所在区域,叉积判断左右 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; const int N=1005; int T,n,m,x1,y1,x2,y2,ans[N],cnt[N]; struct dian { double x,y; dian(double X=0,d…

接替关键:和上题类似,输出不同,注意输入这道题需要排序. #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #include<iostream> using namespace std; typedef long long ll; struct point{ int x,y; }; int n,m,x1,x2…