http://acm.hdu.edu.cn/showproblem.php?pid=3622
对半径进行二分，如果冲突则加入边（按照X AND Y == 0的法则插入边），再2-SAT即可。
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<iostream> 4 #include<cstring> 5 #include<string> 6 #include<cmath> 7 #include<set> 8 #include<list> 9 #include<map> 10 #include<iterator> 11 #include<cstdlib> 12 #include<vector> 13 #include<queue> 14 #include<stack> 15 #include<algorithm> 16 #include<functional> 17 using namespace std; 18 typedef long long LL; 19 #define ROUND(x) round(x) 20 #define FLOOR(x) floor(x) 21 #define CEIL(x) ceil(x) 22 const int maxn=210; 23 const int inf=0x3f3f3f3f; 24 const LL inf64=0x3f3f3f3f3f3f3f3fLL; 25 const double INF=1e30; 26 const double eps=1e-6; 27 int N; 28 struct Point 29 { 30 double x,y; 31 }p[maxn]; 32 struct TwoSAT 33 { 34 int n; 35 vector<int> G[maxn*2]; 36 bool mark[maxn*2]; 37 int S[maxn*2], c; 38 bool dfs(int x) 39 { 40 if (mark[x^1]) return false; 41 if (mark[x]) return true; 42 mark[x] = true; 43 S[c++] = x; 44 for (int i = 0; i < G[x].size(); i++) 45 if (!dfs(G[x][i])) return false; 46 return true; 47 } 48 void init(int n) 49 { 50 this->n = n; 51 for (int i = 0; i < n*2; i++) G[i].clear(); 52 memset(mark, 0, sizeof(mark)); 53 } 54 void add_and_zero(int x,int y) 55 { 56 G[x].push_back(y^1); 57 G[y].push_back(x^1); 58 } 59 bool solve() 60 { 61 for(int i = 0; i < n*2; i += 2) 62 if(!mark[i] && !mark[i+1]) 63 { 64 c = 0; 65 if(!dfs(i)) 66 { 67 while(c > 0) mark[S[--c]] = false; 68 if(!dfs(i+1)) return false; 69 } 70 } 71 return true; 72 } 73 }; 74 TwoSAT sat; 75 double dist(Point A,Point B) 76 { 77 return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)); 78 } 79 void init() 80 { 81 sat.init(N); 82 } 83 void input() 84 { 85 for(int i=0;i<N;i++) 86 scanf("%lf%lf%lf%lf",&p[2*i].x,&p[2*i].y,&p[2*i+1].x,&p[2*i+1].y); 87 } 88 void solve() 89 { 90 double L=0.0,R=40000.0; 91 while(fabs(R-L)>=eps) 92 { 93 double M=(L+R)/2; 94 sat.init(N); 95 for(int i=0;i<2*N-2;i++) 96 for(int j=(i%2==0)? i+2:i+1;j<2*N;j++) 97 if(dist(p[i],p[j])<M) sat.add_and_zero(i,j); 98 if(sat.solve()) L=M; 99 else R=M; 100 } 101 printf("%.2f ",R/2); 102 } 103 int main() 104 { 105 // freopen("in.cpp","r",stdin); 106 while(~scanf("%d",&N)) 107 { 108 init(); 109 input(); 110 solve(); 111 } 112 return 0; 113 }