題目大意：有N對點，給定這N對點的座標，現在要求找出一個最大的半徑R，使得可以從每對點中選擇一個點，並且這N個點以自己為圓心，半徑為R的圓兩兩不相交。

       2-SAT問題解法：對於每一對點，我們且稱之為“兄弟點”；對於這2*N個點，如果兩個點之間“矛盾”，對於本題即兩個圓相交，則連一條邊。對這個圖求強連通分量，然後看是否有某對點屬於同一個強連通分量。如果N對點每一對都屬於不同的強連通分量，滿足。否則不滿足。然後二分半徑即可。

#include <cstdio> #include <cstring> #include <cmath> #define MAXN 210 #define EPS 1e-3 using namespace std; int n; struct Edge{ int v,next; Edge(){} Edge(int _v,int _next):v(_v),next(_next){} }edge[MAXN*MAXN]; int head[MAXN], size; inline void initial(){ memset(head,-1,sizeof(int)*(n+2)); size = 0; } inline void add_edge(int u,int v){ edge[size] = Edge(v,head[u]); head[u] = size ++; } int belong[MAXN]; bool vis[MAXN]; int depth; int bcnt; int dfn[MAXN], low[MAXN]; int stack[MAXN], tail ; inline void Tarjan(int u){ dfn[u] = low[u] = depth ++; vis[u] = 1; stack[++tail] = u; for(int i = head[u];i != -1;i = edge[i].next){ int v = edge[i].v; if(dfn[v] == -1){ Tarjan(v); if(low[u] > low[v]) low[u] = low[v]; }else if(vis[v] && low[u] > dfn[v]){ low[u] = dfn[v]; } } if(low[u] == dfn[u]){ bcnt ++; int j; do{ j = stack[tail--]; vis[j] = 0; belong[j] = bcnt; }while(j != u); } } double x[MAXN], y[MAXN]; inline double Cal(int i,int j){ return (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]); } inline bool can(){ int N = n/2; for(int i = 1;i <= N; ++i) if(belong[i] == belong[i+N]) return false; return true; } int main() { int N; while(scanf("%d",&N) != EOF){ n = 2*N; for(int i = 1;i <= N; ++i) scanf("%lf%lf%lf%lf",&x[i],&y[i],&x[i+N],&y[i+N]); double right = 10000.0*10000.0, left = 0.0, mid; while(right - left >= EPS){ mid = (right + left)/2.0; initial(); for(int i = 1;i < n; ++i) for(int j = i + 1;j <= n; ++j){ //if(i + N == j) continue; if(Cal(i,j) < mid){ int u, v; if(i <= N) u = i + N; else u = i - N; if(j <= N) v = j + N; else v = j - N; add_edge(i,v); add_edge(j,u); } } tail = depth = bcnt = 0; memset(vis,0,sizeof(bool)*(n+2)); memset(dfn,-1,sizeof(int)*(n+2)); memset(low,0,sizeof(int)*(n+2)); for(int i = 1;i <= n; ++i) if(dfn[i] == -1) Tarjan(i); if(can()) left = mid; else right = mid; } printf("%.2lf/n",sqrt(left)/2.0); } }