本题测试点5是从小岛范围内可以直接跳到岸边…… 测试点4是验证步数第一跳最小的情况，刚开始没有考虑回溯，所以错了……
;const int inf=1<<30;struct node{double x,y;} a[100+5];int n,vis[100+5],b[100+5],ans[100+5],cnt;double d,ansfirst,nowfirst;int dis(node d1,node d2){if(d*d<(d1.x-d2.x)*(d1.x-d2.x)+(d1.y-d2.y)*(d1.y-d2.y)) return 0;return 1;}int first(int d1){if(sqrt(a[d1].x*a[d1].x+a[d1].y*a[d1].y)>d+7.5) return 0;;}double first1(int d1){return sqrt(a[d1].x*a[d1].x+a[d1].y*a[d1].y)-7.5;}int safe(node d1){if(d1.x>=50-d) return 1;if(d1.y>=50-d) return 1;if(d1.x<=-50+d) return 1;if(d1.y<=-50+d) return 1;return 0;}void dfs(int d1,int now){int i;if(safe(a[d1])){//printf(“%d %.2f\n”,now,nowfirst);if(now<cnt){for(i=0; i<now; i++)ans[i]=b[i];cnt=now;ansfirst=nowfirst;}else if(now==cnt&&ansfirst>nowfirst){for(i=0; i<now; i++)ans[i]=b[i];cnt=now;ansfirst=nowfirst;}return ;}else{for(i=1; i<=n; i++){if(!vis[i]&&dis(a[d1],a[i])){vis[i]=1;b[now]=i;dfs(i,now+1);vis[i]=0;}}}return;}int main(){int i;while(~scanf(“%d%lf”,&n,&d)){a[0].x=a[0].y=0;for(i=1; i<=n; i++){scanf(“%lf%lf”,&a[i].x,&a[i].y);}if(d+7.5>=50){printf(“1\n”);return 0;}ansfirst=(double)inf;cnt=inf;memset(ans,0,sizeof(ans));for(i=1; i<=n; i++){memset(vis,0,sizeof(vis));if(!vis[i]&&first(i)){nowfirst=first1(i);if(safe(a[i])){if(ansfirst>nowfirst){ans[0]=i;cnt=1;ansfirst=nowfirst;}}vis[i]=1;memset(b,0,sizeof(b));b[0]=i;dfs(i,1);}}if(cnt==inf) printf(“0\n”);else{printf(“%d\n”,cnt+1);for(i=0; i<cnt; i++){printf(“%.0f %.0f\n”,a[ans[i]].x,a[ans[i]].y);}}}return 0;}
，我要准备好行李启程了，谢谢关心我的家人和朋友，
