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
- oracle语法练习汇总
全是自己一个一个敲出来的啊 啊 啊 --(1)查询20号部门的所有员工信息. --(2)查询所有工种为CLERK的员工的工号.员工名和部门名. select e.empno,e.ename,d.dna ...
- Python_pip_02_利用pip安装模块(以安装pyperclip为例)
>任务:利用pip安装pyperclip模块 >前提 你已经在你的电脑里面安装啦Python2.7的Windows版本,并且已经配置了环境变量 >实现步骤 >>打开你的P ...
- 主键primary key和唯一索引unique index
1)主键一定是唯一性索引,唯一性索引并不一定就是主键. 2)主键就是能够唯一标识表中某一行的属性或属性组,一个表只能有一个主键,但可以有多个候选索引. 3)主键常常与外键构成参照完整性约束,防止出现数 ...
- java中public static void main(String[] args)中String[] args代表什么意思?
这是java程序的入口地址,java虚拟机运行程序的时候首先找的就是main方法.跟C语言里面的main()函数的作用是一样的.只有有main()方法的java程序才能够被java虚拟机欲行,可理解为 ...
- Luogu 1484 种树
Luogu 1792 算是双倍经验. 我们考虑对于一个点,我们要么选它,要么选它周围的两个点. 所以我们考虑用一个堆来维护,每次从堆顶取出最大值之后我们把它的权值记为:它左边的权值加上它右边的权值减去 ...
- Shell表达式,如${file##*/}
Shell表达式,如${file##*/} 2017年10月26日 15:24:40 阅读数:343 今天看一个脚本文件的时候有一些地方不太懂,找了一篇文章看了一些,觉得不错,保留下来. 假设我们定义 ...
- CodeForces 658C Bear and Forgotten Tree 3 (构造)
题意:构造出一个 n 个结点,直径为 m,高度为 h 的树. 析:先构造高度,然后再构造直径,都全了,多余的边放到叶子上,注意直径为1的情况. 代码如下: #pragma comment(linker ...
- 基于XML的DI
三.集合属性注入(包含:为数组注入值.为List注入值.为Set注入值.为Map注入值.为Properties注入值) 集合类定义如下: xml定义如下:仔细看 下面是执行代码: 四.对于 ...
- C# GDI
绘制实心矩形 using (Graphics gp = Graphics.FromImage(bmBlank)) { //... ; Rectangle rec = , y, , );//画一个白块, ...
- linux配置环境变量 - 认识
环境 ubuntu17.04 终端(就是黑框) 认识 环境变量:$PATH 在 ×××/bin 下的命令,可以不用到指定目录下, 比如:安装hbase后,hbase提供一些shell命令在habse ...