/*二分+2-sat题意：在一个二维平面上给你n个炸弹，和2*n个位置，每一行的两个位置仅仅能有一个放炸弹如今炸弹爆炸有一个半径。当炸弹爆炸时两个炸弹的半径化成的圆不能相交，求最大半径二分半径。每次假设一个炸弹可放的两个位置中的一个与其它位置有矛盾，就进行建边。最后推断是否存在这样一组解就可以。*/#include
     
      #include
      
       #include
       
        #define eps 1e-5#define N 210struct node {int u,v,next;}bian[N*N];int head[N],dfn[N],low[N],stac[N],yong,index,top,vis[N],ans,belong[N];void init() {memset(head,-1,sizeof(head));memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));memset(stac,0,sizeof(stac));yong=0;index=0;top=0;;memset(vis,0,sizeof(vis));ans=0;memset(belong,0,sizeof(belong));}void addedge(int u,int v) {bian[yong].u=u;bian[yong].v=v;bian[yong].next=head[u];head[u]=yong++;}int Min(int v,int vv) {return v>vv?vv:v;}void tarjan(int u) {  dfn[u]=low[u]=++index;  stac[++top]=u;  vis[u]=1;  int i;  for(i=head[u];i!=-1;i=bian[i].next) {    int v=bian[i].v;    if(!dfn[v]) {        tarjan(v);        low[u]=Min(low[u],low[v]);    }    else        if(vis[v])        low[u]=Min(low[u],dfn[v]);  }  if(dfn[u]==low[u]) {    ans++;    int t;    do {        t=stac[top--];        belong[t]=ans;        vis[t]=0;    }while(t!=u);  }}struct nodee{int u,v,index;}f[N*N];int n;double distance(nodee d,nodee dd) {return 1.0*(d.u-dd.u)*(d.u-dd.u)+1.0*(d.v-dd.v)*(d.v-dd.v);}int judge(double r) {  int i,j;  init();     for(i=0;i<2*n-1;i++)      for(j=i+1;j<2*n;j++) {        if(i==(j^1))continue;       if(distance(f[i],f[j])<(r*2)*(r*2)) {//假设有冲突就进行        addedge(i,j^1);        addedge(j,i^1);       }      }      for(i=0;i<2*n;i++)        if(!dfn[i])            tarjan(i);    for(i=0;i
        
         eps) {        mid=(left+right)/2;      //  printf("%.2f\n",mid);        if(judge(mid))left=mid+eps;        else            right=mid-eps;      }      printf("%.2f\n",left);   }return 0;}