题目地址：HDU 3622
先二分半径，然后小于该半径的不能选，对这些不能选的点对进行加边。然后判断可行性即可。
代码如下：
#include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL __int64 const int INF=0x3f3f3f3f; const double eqs=1e-3; int head[410], cnt, top, ans, index; int dfn[410], low[410], instack[410], stak[410], belong[410]; struct node { int u, v, next; }edge[1000000]; struct Point { int x, y; }dian[10000]; void add(int u, int v) { edge[cnt].v=v; edge[cnt].next=head[u]; head[u]=cnt++; } void tarjan(int u) { dfn[u]=low[u]=++index; instack[u]=1; stak[++top]=u; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(!dfn[v]) { tarjan(v); low[u]=min(low[u],low[v]); } else if(instack[v]) { low[u]=min(low[u],dfn[v]); } } if(dfn[u]==low[u]) { ans++; while(1) { int v=stak[top--]; instack[v]=0; belong[v]=ans; if(u==v) break; } } } void init() { memset(head,-1,sizeof(head)); cnt=0; memset(dfn,0,sizeof(dfn)); memset(instack,0,sizeof(instack)); top=ans=index=0; } double dist(Point x, Point y) { return sqrt((x.x-y.x)*(x.x-y.x)*1.0+(x.y-y.y)*(x.y-y.y)); } int solve(double mid, int n) { int i, j; init(); for(i=0;i eqs) { mid=(l+r)/2; //printf("%.2lf\n",mid); if(solve(mid, n)) { ans=mid; l=mid; } else r=mid; } printf("%.2lf\n",ans/2.0); } return 0; }