06-图2 Saving James Bond - Easy Version (25 分)
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.
Sample Input 1:
14 20
25 -15
-25 28
8 49
29 15
-35 -2
5 28
27 -29
-8 -28
-20 -35
-25 -20
-13 29
-30 15
-35 40
12 12
Sample Output 1:
Yes
Sample Input 2:
4 13
-12 12
12 12
-12 -12
12 -12
Sample Output 2:
No
#include<cstdio>
#include<cmath>
#include<cstdlib>
const double ISLAND_RADIUS = 15.0 / ;
const double SQUARE_SIZE = 100.0;
const int maxn = ; typedef struct Point{
double x,y;
}Position; Position P[maxn];
int n;
double d;
bool vis[maxn]; void save007();
bool FirstJump(int v);
bool DFS(int v);
bool isSave(int v);
bool Jump(int v1,int v2); int main(){
scanf("%d %lf",&n,&d);
for(int i = ; i < n; i++){
scanf("%lf %lf",&(P[i].x),&(P[i].y));
}
for(int i = ; i < n; i++){
vis[i] = false;
}
save007();
return ;
} void save007(){
bool isVist = false;
for(int i = ; i < n; i++){
if(!vis[i] && FirstJump(i)){
isVist = DFS(i);
if(isVist) break;
}
}
if(isVist) printf("Yes\n");
else printf("No\n");
} bool FirstJump(int v){
return sqrt(P[v].x * P[v].x + P[v].y * P[v].y) <= d + ISLAND_RADIUS;
} bool DFS(int v){
bool answer = false;
vis[v] = true;
if(isSave(v)) return true;
for(int i = ; i < n; i++){
if(!vis[i] && Jump(v,i)){
answer = DFS(i);
}
if(answer) break;
}
return answer;
} bool isSave(int v){
return (abs(P[v].x) >= - d) || (abs(P[v].y) >= - d);
} bool Jump(int v1,int v2){
return sqrt((P[v1].x - P[v2].x)*(P[v1].x - P[v2].x) + (P[v1].y - P[v2].y) * (P[v1].y - P[v2].y)) <= d;
}
06-图2 Saving James Bond - Easy Version (25 分)的更多相关文章
- PTA 06-图2 Saving James Bond - Easy Version (25分)
This time let us consider the situation in the movie "Live and Let Die" in which James Bon ...
- 06-图2 Saving James Bond - Easy Version (25 分)
This time let us consider the situation in the movie "Live and Let Die" in which James Bon ...
- pat05-图2. Saving James Bond - Easy Version (25)
05-图2. Saving James Bond - Easy Version (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作 ...
- 05-图2. Saving James Bond - Easy Version (25)
1 边界和湖心小岛分别算一个节点.连接全部距离小于D的鳄鱼.时间复杂度O(N2) 2 推断每一个连通图的节点中是否包括边界和湖心小岛,是则Yes否则No 3 冗长混乱的函数參数 #include &l ...
- Saving James Bond - Easy Version (MOOC)
06-图2 Saving James Bond - Easy Version (25 分) This time let us consider the situation in the movie & ...
- Saving James Bond - Easy Version 原创 2017年11月23日 13:07:33
06-图2 Saving James Bond - Easy Version(25 分) This time let us consider the situation in the movie &q ...
- PAT Saving James Bond - Easy Version
Saving James Bond - Easy Version This time let us consider the situation in the movie "Live and ...
- PTA 07-图5 Saving James Bond - Hard Version (30分)
07-图5 Saving James Bond - Hard Version (30分) This time let us consider the situation in the movie ...
- 06-图2 Saving James Bond - Easy Version
题目来源:http://pta.patest.cn/pta/test/18/exam/4/question/625 This time let us consider the situation in ...
- 06-图2 Saving James Bond - Easy Version (25 分)
This time let us consider the situation in the movie "Live and Let Die" in which James Bon ...
随机推荐
- ASP.NET MVC3 HtmlHelper用法大全
HTML扩展类的所有方法都有2个参数:以textbox为例子public static string TextBox( this HtmlHelper htmlHelper, string name, ...
- springBoot数据库jpa+对接mybatis
1 spring Data jpa hibernate引领数据访问技术,使用orm对象关系映射来进行数据库访问,通过模型和数据库进行映射,通过操作对象实现对数据库操作,把数据库相关操作从代码中独立出 ...
- Cocos2d-x 网络编程
主要介绍内容:Http协议,Socket协议,webSocket协议, Cocos2d-x中的相关类和方法 1 Http协议 HTTP协议也叫超文本传输协议.是互联网广泛使用的通信协议,常用于B/S架 ...
- Unity3d 脚本与C#Socket服务器传输数据
Test.cs脚本 ------------------------------------------------------------------------------------------ ...
- HDOJ 1164 Eddy's research I
Problem Description Eddy's interest is very extensive, recently he is interested in prime number. Ed ...
- 算法Sedgewick第四版-第1章基础-004一封装交易对象
1. package ADT; /****************************************************************************** * Co ...
- Luogu 3942 将军令
之前写那个(Luogu 2279) [HNOI2003]消防局的设立的时候暴力推了一个树形dp,然后就导致这个题不太会写. 贪心,先把树建出来,然后考虑按照结点深度排个序,每次取出还没有被覆盖掉的深度 ...
- 打印sql语句
root->trace hibernate->trace ,然后,改配置 全文搜索:show_sql,将所有的show_sql改为true. 这样,就会显示sql语句了.
- com.fasterxml.jackson.databind.JavaType.isReferenceType
<dependency> <groupId>org.codehaus.jackson</groupId> <artifactId>jackson-map ...
- C#校验算法列举
以下是工作中常用的几种校验算法,后期将不断更新 和校验 /// <summary> /// CS和校验 /// </summary> /// <param name=&q ...