Ada the Ladybug works as farmer. Its season of cucumbers and she wants to harvest them. There lies many cucumbers all around her house. She wants to choose a direction and follow it until all cucumbers in that direction are collected. Lets consider Ada's house as centerpiece of whole universe, lying on [0,0]. The cucumbers are defined as lines on plane. No cucumber goes through Ada's house (and no cucumber touches it). How many cucumbers can Ada pick in one go if she chooses the best direction possible? Input The first line contains an integer T, the number of test-cases. Each test-case begins with an integer 1 ≤ N ≤ 105 Afterward N lines follow, with four integers -106 ≤ x1,y1,x2,y2 ≤ 106, the beginning and end of each cucumber. Each cucumber has positive length. Sum of all N over all test-cases won't exceed 106 Even though cucumber will not go through house, they might touch, intersect or overlap other cucumbers! Output For each test-case print one integer - the maximal number of cucumbers which could be picked if Ada chooses a direction and picks every cucumber lying in it.
时间: 2024-02-14 08:20:43 浏览: 35
这道题目可以使用计算几何中的极角排序来解决。
首先,将每个黄瓜线段表示成起点和终点之间的向量,然后将它们按照极角排序。这可以通过计算每个向量的极角来完成,然后使用 STL 的 sort 函数进行排序。
接下来,在排序后的向量中,找到一个连续的子序列,使得这些向量的极角在某个范围内。我们可以使用双指针来实现这一步。具体来说,我们可以从极角最小的向量开始,逐个向右移动右指针,直到极角超过了某个范围。然后,我们将左指针向右移动一位,并检查新的子序列是否仍然在范围内。我们重复这个过程,直到右指针到达向量的末尾。
最终,我们可以找到一个包含最多黄瓜线段的子序列,它们都在某个范围内。这些线段在同一方向上,因此可以一次性收集它们。
以下是参考代码:
相关问题
Ada the Ladybug works as farmer. Its season of cucumbers and she wants to harvest them. There lies many cucumbers all around her house. She wants to choose a direction and follow it until all cucumbers in that direction are collected. Lets consider Ada's house as centerpiece of whole universe, lying on [0,0]. The cucumbers are defined as lines on plane. No cucumber goes through Ada's house (and no cucumber touches it). How many cucumbers can Ada pick in one go if she chooses the best direction possible? Input The first line contains an integer T, the number of test-cases. Each test-case begins with an integer 1 ≤ N ≤ 105 Afterward N lines follow, with four integers -106 ≤ x1,y1,x2,y2 ≤ 106, the beginning and end of each cucumber. Each cucumber has positive length. Sum of all N over all test-cases won't exceed 106 Even though cucumber will not go through house, they might touch, intersect or overlap other cucumbers! Output For each test-case print one integer - the maximal number of cucumbers which could be picked if Ada chooses a direction and picks every cucumber lying in it.的代码
以下是参考代码,使用了 C++ STL 中的 sort 函数和双指针来实现:
```c++
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
const double EPS = 1e-9;
struct Vector {
double x, y;
double angle; // 极角
};
// 计算点 P 的极角
double polar_angle(double x, double y) {
return atan2(y, x);
}
// 计算向量 AB 的极角
double polar_angle(double Ax, double Ay, double Bx, double By) {
return atan2(By - Ay, Bx - Ax);
}
// 计算向量 AB 的长度
double length(double Ax, double Ay, double Bx, double By) {
return hypot(Bx - Ax, By - Ay);
}
// 计算向量 AB 和向量 CD 的交点
bool intersection(double Ax, double Ay, double Bx, double By,
double Cx, double Cy, double Dx, double Dy,
double& Px, double& Py) {
double d = (Bx - Ax) * (Dy - Cy) - (By - Ay) * (Dx - Cx);
if (abs(d) < EPS) return false; // 平行或重合
double t = ((Cx - Ax) * (Dy - Cy) - (Cy - Ay) * (Dx - Cx)) / d;
Px = Ax + t * (Bx - Ax);
Py = Ay + t * (By - Ay);
return true;
}
// 将向量按照极角排序
bool cmp(const Vector& a, const Vector& b) {
return a.angle < b.angle;
}
int main() {
int T;
cin >> T;
while (T--) {
int N;
cin >> N;
vector<Vector> cucumbers;
for (int i = 0; i < N; ++i) {
double x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
double len = length(x1, y1, x2, y2);
double angle = polar_angle(x1, y1, x2, y2);
cucumbers.push_back({x1, y1, angle});
cucumbers.push_back({x2, y2, angle + (angle < 0 ? 2 * M_PI : 0)});
}
sort(cucumbers.begin(), cucumbers.end(), cmp);
int ans = 0;
int left = 0, right = 0;
while (left < cucumbers.size()) {
while (right < cucumbers.size() && cucumbers[right].angle - cucumbers[left].angle <= M_PI) {
++right;
}
ans = max(ans, right - left);
++left;
}
cout << ans << endl;
}
return 0;
}
```
这个算法的时间复杂度为 $O(n \log n)$,其中 $n$ 是黄瓜的数量。
用python写一个熊猫坐瓢虫的游戏
好的,下面是一个简单的熊猫坐瓢虫游戏代码示例,需要用到Python的pygame库:
```python
import pygame
import random
# 初始化Pygame
pygame.init()
# 设置屏幕大小和标题
screen_width = 800
screen_height = 600
screen = pygame.display.set_mode((screen_width, screen_height))
pygame.display.set_caption("Panda and Ladybug")
# 加载图片
panda_img = pygame.image.load("panda.png").convert_alpha()
ladybug_img = pygame.image.load("ladybug.png").convert_alpha()
# 设置游戏参数
panda_pos = [screen_width // 2, screen_height // 2] # 熊猫初始位置
ladybug_list = [] # 瓢虫列表
score = 0
# 创建瓢虫类
class Ladybug:
def __init__(self):
self.x = random.randint(0, screen_width - 50)
self.y = random.randint(0, screen_height - 50)
self.speed = 2
def move(self):
direction = random.choice(["up", "down", "left", "right"])
if direction == "up":
self.y -= self.speed
elif direction == "down":
self.y += self.speed
elif direction == "left":
self.x -= self.speed
elif direction == "right":
self.x += self.speed
# 检查瓢虫是否出界
if self.x < 0:
self.x = 0
elif self.x > screen_width - 50:
self.x = screen_width - 50
if self.y < 0:
self.y = 0
elif self.y > screen_height - 50:
self.y = screen_height - 50
def draw(self):
screen.blit(ladybug_img, (self.x, self.y))
# 创建多个瓢虫
for i in range(10):
ladybug = Ladybug()
ladybug_list.append(ladybug)
# 游戏循环
running = True
while running:
# 处理事件
for event in pygame.event.get():
if event.type == pygame.QUIT:
running = False
# 移动瓢虫
for ladybug in ladybug_list:
ladybug.move()
# 检查熊猫是否碰到瓢虫
for ladybug in ladybug_list:
if (panda_pos[0] < ladybug.x + 50 and panda_pos[0] + 50 > ladybug.x and
panda_pos[1] < ladybug.y + 50 and panda_pos[1] + 50 > ladybug.y):
score += 1
ladybug_list.remove(ladybug)
# 移动熊猫
keys = pygame.key.get_pressed()
if keys[pygame.K_UP]:
panda_pos[1] -= 5
elif keys[pygame.K_DOWN]:
panda_pos[1] += 5
elif keys[pygame.K_LEFT]:
panda_pos[0] -= 5
elif keys[pygame.K_RIGHT]:
panda_pos[0] += 5
# 检查熊猫是否出界
if panda_pos[0] < 0:
panda_pos[0] = 0
elif panda_pos[0] > screen_width - 50:
panda_pos[0] = screen_width - 50
if panda_pos[1] < 0:
panda_pos[1] = 0
elif panda_pos[1] > screen_height - 50:
panda_pos[1] = screen_height - 50
# 绘制游戏场景
screen.fill((255, 255, 255))
screen.blit(panda_img, tuple(panda_pos))
for ladybug in ladybug_list:
ladybug.draw()
# 绘制分数
font = pygame.font.Font(None, 36)
score_text = font.render("Score: {}".format(score), True, (0, 0, 0))
screen.blit(score_text, (10, 10))
# 更新屏幕
pygame.display.update()
# 退出Pygame
pygame.quit()
```
在代码中,我们创建了一个Ladybug类表示瓢虫,它包含瓢虫的位置和移动方法。我们创建了多个瓢虫,并在游戏循环中移动它们。我们还检查熊猫是否碰到了瓢虫,并根据情况增加得分。最后我们在屏幕上绘制了游戏场景和分数,并更新屏幕。
相关推荐
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![gz](https://img-home.csdnimg.cn/images/20210720083447.png)
![gz](https://img-home.csdnimg.cn/images/20210720083447.png)