POJ 2398 Toy Storage 二分+叉积
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 rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:
We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.
Input
The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard. A line consisting of a single 0 terminates the input.
Output
For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.
Sample Input
4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0
Sample Output
Box
2: 5
Box
1: 4
2: 1
这题和上一题poj2318 差不多。无非就是点不是顺序来的,排个序就好了,
题目几乎没有差别。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL ;
const int maxn = 5e3 + ;
int ans[maxn], num[maxn];
struct point {
int x, y;
point() {}
point(int x, int y) : x(x), y(y) {}
point operator - (const point &a) const {
return point(x - a.x, y - a.y);
}
} ;
struct line {
point first, second;
line() {};
line(point first, point second ): first(first), second(second) {}
} Line[maxn];
bool cmp(line a, line b) {
return a.first.x < b.first.x;
}
int cross(point a, point b) {
return a.x * b.y - a.y * b.x;
}
int main() {
int n, m, upx, upy, downx, downy;
while(scanf("%d", &n), n) {
memset(ans, , sizeof(ans));
memset(num, , sizeof(num));
scanf("%d%d%d%d%d", &m, &upx, &upy, &downx, &downy );
for (int i = ; i < n ; i++ ) {
int x, y;
scanf("%d%d", &x, &y);
Line[i] = line(point(x, upy), point(y, downy));
}
sort(Line,Line+n,cmp);
for (int i = ; i < m ; i++) {
point p;
scanf("%d%d", &p.x, &p.y);
int l = , r = n - , cnt = n, mid;
while(l <= r) {
mid = (l + r) / ;
if (cross(Line[mid].first - p, Line[mid].second - p) <= ) {
r = mid - ;
cnt = mid;
} else l = mid + ;
}
ans[cnt]++;
}
for (int i = ; i <= n ; i++) {
if (ans[i]) num[ans[i]]++;
}
printf("Box\n");
for (int i = ; i <= n ; i++) {
if (num[i]) printf("%d: %d\n", i, num[i]);
}
}
return ;
}
POJ 2398 Toy Storage 二分+叉积的更多相关文章
- POJ 2398 Toy Storage (叉积判断点和线段的关系)
题目链接 Toy Storage Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4104 Accepted: 2433 ...
- POJ 2398 Toy Storage(叉积+二分)
Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finis ...
- poj 2398 Toy Storage(计算几何)
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 ...
- POJ 2318 TOYS && POJ 2398 Toy Storage(几何)
2318 TOYS 2398 Toy Storage 题意 : 给你n块板的坐标,m个玩具的具体坐标,2318中板是有序的,而2398无序需要自己排序,2318要求输出的是每个区间内的玩具数,而231 ...
- 向量的叉积 POJ 2318 TOYS & POJ 2398 Toy Storage
POJ 2318: 题目大意:给定一个盒子的左上角和右下角坐标,然后给n条线,可以将盒子分成n+1个部分,再给m个点,问每个区域内有多少各点 这个题用到关键的一步就是向量的叉积,假设一个点m在 由ab ...
- 2018.07.04 POJ 2398 Toy Storage(二分+简单计算几何)
Toy Storage Time Limit: 1000MS Memory Limit: 65536K Description Mom and dad have a problem: their ch ...
- poj 2318 TOYS & poj 2398 Toy Storage (叉积)
链接:poj 2318 题意:有一个矩形盒子,盒子里有一些木块线段.而且这些线段坐标是依照顺序给出的. 有n条线段,把盒子分层了n+1个区域,然后有m个玩具.这m个玩具的坐标是已知的,问最后每一个区域 ...
- POJ 2398 Toy Storage(计算几何,叉积判断点和线段的关系)
Toy Storage Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3146 Accepted: 1798 Descr ...
- POJ 2398 - Toy Storage 点与直线位置关系
Toy Storage Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5439 Accepted: 3234 Descr ...
随机推荐
- react ant-design自定义图标
ant-design给我们提供的图标不够怎么办呢?答案是我们可以自定义图标. 自定义图标也挺简单的,现在图标推荐用svg格式,那么我们就需要制作svg图片. 下面让我们看看如果制作svg图片吧. 1. ...
- js 实现字符串转日期进行比较大小
代码如下 var a = '2016-01-01 12:12:12'; var b = '2016-01-01 12:12:13'; var al = new Date(a).getTime(); v ...
- express与ejs,ejs在Linux上面的路径问题
1.学习使用ejs模板(这个是ejs.js) var express = require('express'); var app = express(); app.set("view eng ...
- Mysql数据库的压力
rationalError: (2006, 'MySQL server has gone away') 2017年10月10日 20:04:43 阅读数:377 问题描述 使用django+celer ...
- 30分钟 带你浅入seajs源码
上个星期写了浅入requirejs的, 大家都知道 require是AMD规范(Asynchronous Module Definition) 来 今天我们一起看看 CMD规范(Common Mo ...
- Redis进阶:数据持久化,安全,在PHP中使用
一.redis数据持久化 由于redis是一个内存数据库,如果系统遇到致命问题需要关机或重启,内存中的数据就会丢失,这是生产环境所不能允许的.所以redis提供了数据持久化的能力. redis提供了两 ...
- IDEA 中.properties文件中文自动转Unicode编码及乱码问题
问题描述: 在使用IDEA开发工具编辑属性文件(.properties)的时候出现中文自动转成了Unicode编码,或在读取属性文件的时候中文出现乱码. 问题解决: 进入 File -> Set ...
- 使用JDK自带的keytool工具生成证书
一.keytool 简介 keytool 是java用于管理密钥和证书的工具,它使用户能够管理自己的公钥/私钥对及相关证书,用于(通过数字签名)自我认证(用户向别的用户/服务认证自己)或数据完整性以及 ...
- Assetbundle1
AssetBundle运行时加载:来自文件就用CreateFromFile(注意这种方法只能用于standalone程序)这是最快的加载方法也可以来自Memory,用CreateFromMemory( ...
- 用tensorflow实现自然语言处理——基于循环神经网络的神经语言模型
自然语言处理和图像处理不同,作为人类抽象出来的高级表达形式,它和图像.声音不同,图像和声音十分直觉,比如图像的像素的颜色表达可以直接量化成数字输入到神经网络中,当然如果是经过压缩的格式jpeg等必须还 ...