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.

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

分析

大概意思就是 007 站在边长为 100 的矩形水池中间直径为 15 的小岛上，他所处位置是 (0,0)，最右上角的位置是 (50,50)，池塘中间不同位置有鳄鱼，他能否踩着鳄鱼上岸
考察图的遍历
我的办法是记录每只鳄鱼的信息：

• 该鳄鱼的横纵坐标
  方便计算距离
• 该鳄鱼能否最开始一步跳上
  对应不同的连通图
• 能否从该鳄鱼上岸
  结束条件

记录下上述信息后遍历全部能"开始一步跳上"的鳄鱼，即每个连通图，如果遇到某个鳄鱼 “能上岸”，即退出遍历，输出“Yes”，如果全部连通图都遍历完成，还是不能上岸，则输出“No”
其中遍历分别用了 DFS 和 BFS 实现

#include<iostream>
#include<stdlib.h>
#include<cmath>
#include<queue> 
#define MaxVertex 105
struct Node{  // 存鳄鱼信息
	int hor;   // 横坐标 
	int ver;  // 纵坐标
	bool visit;  // 是否被访问
	bool safe;  // 是否能上岸 
	bool jump;  // 第一步能否跳上去 
};
int N;   // 鳄鱼数 
int D;   // 跳跃距离
bool isSafe;  // 是否上岸 
Node G[MaxVertex];
using namespace std;
const double diameter=15;

// 计算两点距离 
double getLen(int x1,int y1,int x2,int y2){
	return sqrt(pow(x1-x2,2.0)+pow(y1-y2,2.0));
}

// 计算该鳄鱼能否到岸边 
bool ashore(int x,int y){
	// 分别计算当前结点与岸边的距离
	// 即与 (x,50),(x,-50),(50,y),(-50,y) 的距离 
	if(abs(x-50)<=D || abs(x+50)<=D || abs(y+50)<=D || abs(y-50)<=D)
		return true;
	return false;
}

// 确认鳄鱼是否安全("能上岸") 
void getSafe(){
	for(int i=0;i<N;i++){
		// 如果该鳄鱼位置和"岸边"相邻 
		if(ashore((G[i].hor),(G[i].ver)))
			G[i].safe = true; // 将情况置为 true
		else
			G[i].safe = false; 
	}
}

// 确认哪些鳄鱼是可以第一步跳上去的 
void getJump(){
	for(int i=0;i<N;i++){
		// 如果该鳄鱼位置和"湖中心"相邻（跳跃距离+半径） 
		if(getLen(G[i].hor,G[i].ver,0,0)<=D+diameter/2)
			G[i].jump = true;
		else
			G[i].jump = false;
	}
}

// 初始化 
void Init(){
	cin>>N>>D;
	int x,y;
	for(int i=0;i<N;i++){
		cin>>x>>y;
		G[i].hor = x;
		G[i].ver = y;
		G[i].visit = false;
	}
	getSafe();
	getJump();
	isSafe = false;
}
/* void DFS(int v){ if(G[v].safe){ isSafe = true; return; } G[v].visit = true; for(int i=0;i<N;i++){ // 距离如果小于 D，且未跳过，则能跳 if(getLen(G[v].hor,G[v].ver,G[i].hor,G[i].ver)<=D && !G[i].visit) DFS(i); } } */
void BFS(int v){
	queue<Node> q;
	Node tmp;
	G[v].visit = true;
	// 第一只鳄鱼入队 
	q.push(G[v]);
	while(!q.empty()){
		tmp = q.front();
		q.pop();
		// 能上岸 
		if(tmp.safe){
			isSafe = true;
			return;
		}
		for(int i=0;i<N;i++){
			// 距离如果小于 D，且未跳过，则能跳
			if(getLen(tmp.hor,tmp.ver,G[i].hor,G[i].ver)<=D && !G[i].visit){
				G[i].visit = true;
				q.push(G[i]);
			}
		}
	}
}

// 遍历所有第一步能跳到的鳄鱼 
void listCompoent(){
	for(int i=0;i<N;i++)
		if(G[i].jump){
		// DFS(i); 
			BFS(i);
		}
	if(isSafe)
		cout<<"Yes"<<endl;
	else
		cout<<"No"<<endl;
}


int main(){
	Init();
	listCompoent();
	return 0;
}