很麻烦的题目
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <stack>
#include <queue>
#include <cstring>
using namespace std;
int E,N;
int visited[200];
int path[200]={-1}; //保存上一个节点
struct crocodile
{int x; //x坐标int y; //y坐标int d; //与0,0距离的平方
}croco[200];
int reachFromCenter(crocodile p){int a=(7.5+E)*(7.5+E);if(p.d<=a) return p.d;else return 0;
}
int cmp(const void *a,const void *b){return ((*(crocodile *)a).d)-((*(crocodile *)b).d);
}
int success(crocodile a){return a.x>=50-E||a.y>=50-E||a.x-E<=-50||a.y-E<=-50;
}
void init(){memset(visited,-1,sizeof(visited));
}
int calDist(crocodile a,crocodile b){ //计算两点之间的距离return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int main(){cin>>N>>E;for(int i=0;i<=N-1;i++){cin>>croco[i].x>>croco[i].y;croco[i].d=croco[i].x*croco[i].x+croco[i].y*croco[i].y; //计算与0，0的距离}qsort(croco,N,sizeof(croco[0]),cmp); //按照距离排一次序if(E>=42.5){printf("1\n"); //一跳就能跳出，输出一，没踩鳄鱼，所有没有鳄鱼return 0;}else{ //必须要踩鳄鱼了queue <int> q; //bfs必备的队列 int last;for(int i=0;i<=N-1;i++){init();if(reachFromCenter(croco[i])){ //所有第一步能跳的进队q.push(i);visited[i]=0;last=i;}}int step=2;int tail;while(!q.empty()){int element=q.front();q.pop();if(success(croco[element])){int k=1;stack <int> s;printf("%d\n",step);while(k<step){k++;s.push(element); //从最后一条鳄鱼到第一条鱼开始压栈element=path[element]; //上一条鳄鱼}while(!s.empty()){element=s.top(); //出栈s.pop();printf("%d %d\n",croco[element].x, croco[element].y);}return 0; //直接跳出}for(int i=0;i<=N-1;i++){if(visited[i]==-1&&calDist(croco[element],croco[i])<=E*E){q.push(i);path[i]=element;visited[i]=0;tail=i;}}if(last==element){step++;last=tail;}}if(q.empty()){ //跳完了所有鳄鱼还跳不出只能死在岛上printf("0\n");}}return 0;
}