# 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 NN (\le 100≤100), the number of crocodiles, and DD, the maximum distance that James could jump. Then NN lines follow, each containing the (x, y)(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 <iostream>
#include <cmath>
#include <vector>
using namespace std;

class Crocodile{
private:
int x;                  //x坐标
int y;                  //y坐标
public:
//构造函数
Crocodile(){
this->x = 0;
this->y = 0;
}

Crocodile(int x,int y){
this->x = x;
this->y = y;
}

int getX(){
return this->x;
}

int getY(){
return this->y;
}
};

class Save007{
private:
vector<Crocodile> crocodiles;
int* isVisited;
int MaxdDistance;
int Maxnum;
public:
Save007(int maxnum,int maxdistance){
this->Maxnum = maxnum;
this->isVisited = new int[maxnum];
this->MaxdDistance = maxdistance;
for ( int i = 0 ; i < this->Maxnum ; i++){
int x,y;
cin>>x>>y;
Crocodile tmp(x,y);
this->crocodiles.push_back(tmp);
this->isVisited[i] = 0;
}
}

//判断该鳄鱼是否能让007跳上岸
int Judge(Crocodile crocodile){
int distX = 50 - abs(crocodile.getX());
int distY = 50 - abs(crocodile.getY());

if ( distX <= this->MaxdDistance || distY <= this->MaxdDistance){
return 1;
}else{
return 0;
}
}

//计算007是否能跳到这只鳄鱼上
int Caculate(Crocodile crocodile1,Crocodile crocodile2,double jump_distance){
int distX =  crocodile1.getX() - crocodile2.getX();
int distY =  crocodile1.getY() - crocodile2.getY();
double dist = sqrt(pow(distX,2)+ pow(distY,2));
if (dist <= jump_distance){
return 1;
}else{
return 0;
}
}

//DFS遍历
void DFS(int start){
int flag  = 0;
this->isVisited[start] = 1;
if (this->Judge(this->crocodiles[start])){
cout<<"Yes";
exit(0);
}else{
for ( int i = 0 ; i < this->Maxnum ; i++){
int judge = this->Caculate(this->crocodiles[start],this->crocodiles[i],this->MaxdDistance);
if ( !this->isVisited[i] && judge){
DFS(i);
}
}
}
}

void Save(){
int start;
int result = 0;
vector<int> first_jump;
//第一跳单独拿出来
for ( int i = 0 ; i < this->Maxnum ; i++){
Crocodile tmp(0,0);
if (this->Caculate(this->crocodiles[i], tmp ,this->MaxdDistance+7.5)){
first_jump.push_back(i);
}
}

if ( first_jump.empty()){
cout<<"No";
return;
}else{
for (int i = 0 ; i < first_jump.size() ; i++){
DFS(first_jump[i]);
}
cout<<"No";
return;
}
}
};

int main()
{
int maxnum,maxdistance;
cin>>maxnum>>maxdistance;
Save007 save(maxnum,maxdistance);

save.Save();

return 0;
} ```

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