Saving James Bond（c++）

题目描述：

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.

输入1

输出1

Yes

输入2

输出2

No

解题代码

// 深搜
#include<bits/stdc++.h>
using namespace std;
int visited[100],n,d;
struct coordinate{      //结构体存放鳄鱼的位置 int x;int y;
}crocodile[105];    //每次是否能跳跃的函数
bool jump(int i,int j)
{return (sqrt(pow(crocodile[i].x-crocodile[j].x,2)+pow(crocodile[i].y-crocodile[j].y,2))<=d);
}//第一次跳跃的函数，因为中间是小岛所以跳跃长度需要加上小岛的直径
bool firstjump(int i)
{return (sqrt(pow(crocodile[i].x,2)+pow(crocodile[i].y,2))<=(d+15));
}bool isarrive(int i)
{int x_length,y_length;x_length=abs(crocodile[i].x)-50;    //看最后一只鳄鱼的位置距离岸的位置 y_length=abs(crocodile[i].y)-50;    //看最后一只鳄鱼的位置距离岸的位置//看看横纵距离是否都小于能跳到的距离，是就说明有答案 if(abs(x_length)<=d || abs(y_length)<=d) return true;else return false;} //深搜
int dfs(int i)
{int ans=0;if(isarrive(i)) //如果上岸了 ,标定 ans=1;else         //没上岸继续搜 {visited[i]=true;for(int j=0;j<n;j++){if(!visited[j]&&jump(i,j))    //搜索条件：还没走过并且跳跃距离之内 {ans=dfs(j);if(ans==1) break; //找到答案就退出寻找 }   }} return ans;
}int main()
{int ans=0;cin>>n>>d;for(int i=0;i<n;i++)cin>>crocodile[i].x>>crocodile[i].y;memset(visited,false,sizeof(visited));  //初始化，让每个位置都是没走过 //遍历所有鳄鱼点，判断，搜索 for(int i=0;i<n;i++){if(!visited[i]&&firstjump(i)){ans=dfs(i);if(ans==1){cout<<"Yes";return 0;} }    }cout<<"No";return 0;
}