浙江大学-数据结构-习题-06-图2 Saving James Bond - Easy Version (25 分)

该博客介绍了浙江大学数据结构课程的一个习题,涉及到James Bond如何在被鳄鱼包围的湖中逃脱。问题转化为图的联通性判断,给定鳄鱼数量、最大跳跃距离和鳄鱼坐标,需确定James能否跳到湖岸。博客讨论了解题思路,包括将可达的鳄鱼和湖岸视为连通,并提供了伪代码和C++代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

浙江大学-数据结构-习题-06-图2 Saving James Bond - Easy Version (25 分)

题目

This time let us consider the situation in the movie “Live and Let Die” in which James Bond, the world’s most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape – he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head… Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).
Assume that the lake is a 100 by 100 square one. Assume that the center of the lake is at (0,0) and the northeast corner at (50,50). The central island is a disk centered at (0,0) with the diameter of 15. A number of crocodiles are in the lake at various positions. Given the coordinates of each crocodile and the distance that James could jump, you must tell him whether or not he can escape.
Input Specification:
Each input file contains one test case. Each case starts with a line containing two positive integers N (≤100), the number of crocodiles, and D, the maximum distance that James could jump. Then N lines follow, each containing the (x,y) location of a crocodile. Note that no two crocodiles are staying at the same position.
Output Specification:
For each test case, print in a line “Yes” if James can escape, or “No” if not.

1.大意分析

在一个100*100的湖中央有一个直径15的小岛(独特的大结点),岛外分布有若干个鳄鱼(小结点),007从小岛出发,通过鳄鱼跳出湖。
输入第一行给出鳄鱼的个数N,007的跳跃距离D。
随后各行给出鳄鱼的坐标。

2.思路

将距离不大于D的若干个结点与结点和结点与湖岸之间视为联通,若能有一条联通的路线从中央小岛通往湖岸,则输出"Yes",否则输出"No"。

3.Mooc伪代码:

Mooc算法

4.C++代码:

#include <iostream>
#include <cmath>
using namespace std;

typedef struct node{
   //存储输入的节点
    int x;
    int y;
}Vertex;

int N
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值