http://poj.org/problem?id=2318

第一次完全是$O(n^2)$的暴力为什么被卡了~QAQ(一定是常数太大了...)

后来排序了下点然后单调搞了搞。。(然而还是可以随便造出让我的code变成$O(n^2)$的23333)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mkpii make_pair<int, int>
#define pdi pair<double, int>
#define mkpdi make_pair<double, int>
#define pli pair<ll, int>
#define mkpli make_pair<ll, int>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << '\t'; cout << endl
inline const ll getint() { ll r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const int N=5015;
struct dat { ll x, y; }E[2], a[N][2], b[N];
ll Cross(dat &a, dat &b, dat &c) {
static ll x1, x2, y1, y2;
x1=a.x-c.x; y1=a.y-c.y;
x2=b.x-c.x; y2=b.y-c.y;
return x1*y2-x2*y1;
}
int ans[N]; bool check(dat &l1, dat &l2, dat &r1, dat &r2, dat &p) {
if(Cross(p, l1, l2)*Cross(p, r1, r2)<0) return 1;
return 0;
} int n, m;
void work(int last, dat p) {
int xxx=n+1;
for1(now, last, xxx) {
if(check(a[now-1][0], a[now-1][1], a[now][0], a[now][1], p)) { ++ans[now-1]; return; }
}
} bool cmp(const dat &a, const dat &b) { return a.x<b.x; } int main() {
while(read(n), n) {
read(m);
rep(i, 2) read(E[i].x), read(E[i].y);
a[0][0].x=E[0].x; a[0][0].y=E[0].y;
a[0][1].x=E[0].x; a[0][1].y=E[1].y;
a[n+1][0].x=E[1].x; a[n+1][0].y=E[0].y;
a[n+1][1].x=E[1].x; a[n+1][1].y=E[1].y;
for1(i, 1, n) rep(k, 2) read(a[i][k].x), a[i][k].y=E[k].y;
for1(i, 1, m) read(b[i].x), read(b[i].y);
sort(b+1, b+1+m, cmp);
int last=1;
for1(i, 1, m) {
while(last<=n && max(a[last][0].x, a[last][1].x)<b[i].x) ++last;
work(last, b[i]);
}
rep(i, n+1) printf("%d: %d\n", i, ans[i]), ans[i]=0;
puts("");
}
return 0;
}

  


Description

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

【POJ】2318 TOYS(计算几何基础+暴力)的更多相关文章

  1. POJ 2318 - TOYS - [计算几何基础题]

    题目链接:http://poj.org/problem?id=2318 Time Limit: 2000MS Memory Limit: 65536K Description Calculate th ...

  2. POJ 2318 TOYS(叉积+二分)

    题目传送门:POJ 2318 TOYS Description Calculate the number of toys that land in each bin of a partitioned ...

  3. poj 2318 TOYS (二分+叉积)

    http://poj.org/problem?id=2318 TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 101 ...

  4. 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 ...

  5. POJ 2318 TOYS【叉积+二分】

    今天开始学习计算几何,百度了两篇文章,与君共勉! 计算几何入门题推荐 计算几何基础知识 题意:有一个盒子,被n块木板分成n+1个区域,每个木板从左到右出现,并且不交叉. 有m个玩具(可以看成点)放在这 ...

  6. 简单几何(点与线段的位置) POJ 2318 TOYS && POJ 2398 Toy Storage

    题目传送门 题意:POJ 2318 有一个长方形,用线段划分若干区域,给若干个点,问每个区域点的分布情况 分析:点和线段的位置判断可以用叉积判断.给的线段是排好序的,但是点是无序的,所以可以用二分优化 ...

  7. 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 ...

  8. poj 2318 TOYS &amp; 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 ...

  9. 【POJ】2653 Pick-up sticks(计算几何基础+暴力)

    http://poj.org/problem?id=2653 我很好奇为什么这样$O(n^2)$的暴力能过.... 虽然说这是加了链表优化的,但是最坏不也是$O(n^2)$吗...(只能说数据太弱.. ...

随机推荐

  1. ZBT的计算几何模板

    Basic template 一个基础型模板包括一个向量的实现 DATE: 2015-06-01 #define op operator #define __ while #define _0 ret ...

  2. 【Redis】Redis分布式集群几点说道

    Redis数据量日益增大,使用的公司越来越多,不仅用于做缓存,同时趋向于存储这一块,这样必促使集群的发展,各个公司也在收集适合自己的集群方案,目前行业用的比较多的是下面几种集群架构,大部分都是采用分片 ...

  3. Linux下提取IP至文件

    ifconfig | grep 'inet[^6]' | sed 's/^\s*//g' | cut -d ' ' -f2 > ips.txt 排除127开头的IP: ifconfig | gr ...

  4. jquery 常用类别选择器

    1.$('#showDiv'):  id选择器,相当于javascript中的documentgetElementById("showDiv"); 2.$("onecla ...

  5. iOS 一个工程中引用其他工程时编译的Architecture问题

    当引用了其他工程时,在编译时报错,提示你编译指令架构不对,你需要查看一下这几个工程的Architecture部分是否又冲突,比如主工程设置Valid Architecture为armv7 而 另一个子 ...

  6. php中GD库的简单使用

    在php中需要图像处理的地方GD库会发挥重要的作用,php可以创建并处理包括GIF,PNG,JPEG,WBMP以及XPM在内的多种图像格式,简单的举几个例子: 1.用GD库会创建一块空白图片,然后绘制 ...

  7. Android开发数据库三层应用-DataSnap

    Android开发数据库三层应用-DataSnap http://www.2ccc.com/news/Html/?1517.html 核心提示:我觉得Delphi最强大的的功能之一就是开发数据库三层应 ...

  8. angularjs 指令(directive)详解(1)

    原文地址 什么是directive?我们先来看一下官方的解释: At a high level, directives are markers on a DOM element (such as an ...

  9. sublime 实用 快捷键

    alt+- 向后导航 alt+shift+- 向前导航 ctrl+shift+↑↓ 上下移动一行 ctrl+k,ctrl+u 转换所选为大写 ctrl+k,ctrl+l(字母L) 转换所选为小写 ct ...

  10. https_request

    <?php $access_token = );         curl_setopt($curl, CURLOPT_POSTFIELDS, $data);     }     curl_se ...