poj 2318 TOYS(计算几何 点与线段的关系)
2015-08-17 09:01
513 查看
[b]TOYS[/b]
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 12015 | Accepted: 5792 |
Calculate the number of toys that land in each bin of a partitioned toy box.
Mom and dad have a problem - their child John never puts his toys
away when he is finished playing with them. They gave John a rectangular
box to put his toys in, but John is rebellious and obeys his parents by
simply throwing his toys into the box. All the toys get mixed up, and
it is impossible for John to find his favorite toys.
John's parents came up with the following idea. They put cardboard
partitions into the box. Even if John keeps throwing his toys into the
box, at least toys that get thrown into different bins stay separated.
The following diagram shows a top view of an example toy box.
For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.
Input
The
input file contains one or more problems. The first line of a problem
consists of six integers, n m x1 y1 x2 y2. The number of cardboard
partitions is n (0 < n <= 5000) and the number of toys is m (0
< m <= 5000). The coordinates of the upper-left corner and the
lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The
following n lines contain two integers per line, Ui Li, indicating that
the ends of the i-th cardboard partition is at the coordinates (Ui,y1)
and (Li,y2). You may assume that the cardboard partitions do not
intersect each other and that they are specified in sorted order from
left to right. The next m lines contain two integers per line, Xj Yj
specifying where the j-th toy has landed in the box. The order of the
toy locations is random. You may assume that no toy will land exactly on
a cardboard partition or outside the boundary of the box. The input is
terminated by a line consisting of a single 0.
Output
The
output for each problem will be one line for each separate bin in the
toy box. For each bin, print its bin number, followed by a colon and one
space, followed by the number of toys thrown into that bin. Bins are
numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate
the output of different problems by a single blank line.
Sample Input
5 6 0 10 60 0 3 1 4 3 6 8 10 10 15 30 1 5 2 1 2 8 5 5 40 10 7 9 4 10 0 10 100 0 20 20 40 40 60 60 80 80 5 10 15 10 25 10 35 10 45 10 55 10 65 10 75 10 85 10 95 10 0
Sample Output
0: 2 1: 1 2: 1 3: 1 4: 0 5: 1 0: 2 1: 2 2: 2 3: 2 4: 2
Hint
As the example illustrates, toys that fall on the boundary of the box are "in" the box.
Source
Rocky Mountain 2003
开始正经的学计算几何,恩,是的没错~
加油~
题意:给定一个长方形,在里面加上不相交的线,然后给若干点,求这些点落在哪个区域。
#include <iostream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <math.h> #include <algorithm> #include <cctype> #include <string> #include <map> #define N 500015 #define INF 1000000 #define ll long long using namespace std; struct Point { int x,y; Point(){} Point(int _x,int _y) { x = _x;y = _y; } Point operator -(const Point &b)const { return Point(x - b.x,y - b.y); } int operator *(const Point &b)const { return x*b.x + y*b.y; } int operator ^(const Point &b)const { return x*b.y - y*b.x; } }; struct Line { Point s,e; Line(){} Line(Point _s,Point _e) { s = _s;e = _e; } }; int xmult(Point p0,Point p1,Point p2) //计算p0p1 X p0p2 { return (p1-p0)^(p2-p0); } const int MAXN = 5050; Line line[MAXN]; int ans[MAXN]; int main(void) { int n,m,x1,y1,x2,y2,i; int ui,li; int cnt = 0; while(scanf("%d",&n),n) { if(cnt == 0) cnt = 1; else printf("\n"); scanf("%d %d %d %d %d",&m,&x1,&y1,&x2,&y2); for(i = 0; i < n; i++) { scanf("%d%d",&ui,&li); line[i] = Line(Point(ui,y1),Point(li,y2)); } line = Line(Point(x2,y1),Point(x2,y2)); int x,y; Point p; memset(ans,0,sizeof(ans)); while(m--) { scanf("%d %d",&x,&y); p = Point(x,y); int l = 0,r = n,tmp = 0; while(l <= r) { int mid = (l + r)/2; if(xmult(p,line[mid].s,line[mid].e) < 0) { tmp = mid; r = mid - 1; } else l = mid + 1; } ans[tmp]++; } for(i = 0; i <= n; i++) printf("%d: %d\n",i,ans[i]); } return 0; }
相关文章推荐
- phpStorm快捷键
- 一周 GNOME 之旅:品味它和 KDE 的是是非非(第一节 介绍)
- 11 函数和函数式编程 - 《Python 核心编程》
- 08-14工作总结
- 利用 Android Studio 和 Gradle 打包多版本APK
- 饭卡
- 通过ApplicationContextAwareSpring实现手工加载配置的javabean
- Tomcat中web.xml的配置
- codeforces#316 div2
- Unity 内的敌人AI
- 饭卡
- 通过ApplicationContextAwareSpring实现手工加载配置的javabean
- PowerShell 自动备份oracle并上传到ftp
- GDOI2016模拟8.16打发时间
- 你把它列入博客设置?
- 【Ajax技术】JQuery处理XML数据
- Nginx服务器限制IP访问的各种情况全解析
- 通达OA 小飞鱼OA实施法:以项目管理的方式来推进工作流设计项目实施
- hdu 5391 Zball in Tina Town(威尔逊定理)
- 设置字符串中某些字符的特殊效果