zoj 3370(二分+二分图染色)

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3912

思路：二分覆盖直径，然后判断是否有冲突（即距离小于等于直径的不能使用同一频率），这样可以用二分图染色的办法判断，看是否能将整个图都染上色。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<cmath>
 using namespace std;
 #define MAXN 1444
 #define inf 1<<30
 #define eps 1e-9

 struct Point{
     double x,y;
 }p[MAXN];

 double map[MAXN][MAXN];
 int color[MAXN];
 int ans[MAXN];
 int n;

 double Get_Dist(int i,int j)
 {
     double d1=(p[i].x-p[j].x)*(p[i].x-p[j].x);
     double d2=(p[i].y-p[j].y)*(p[i].y-p[j].y);
     return sqrt(d1+d2);
 }

 bool Judge(double limit)
 {
     memset(color,,sizeof(color));
     queue<int>que;
     for(int i=;i<=n;i++){
         if(color[i]==){
             color[i]=;
             que.push(i);
             while(!que.empty()){
                 int u=que.front();
                 que.pop();
                 for(int v=;v<=n;v++){
                     if(map[u][v]>limit-eps)continue;
                     if(color[u]==color[v])return false;
                     if(color[v]==){
                         color[v]=-color[u];
                         que.push(v);
                     }
                 }
             }
         }
     }
     for(int i=;i<=n;i++)ans[i]=color[i];
     return true;
 }

 int main()
 {
     while(~scanf("%d",&n)){
         for(int i=;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
         for(int i=;i<=n;i++)
             for(int j=;j<=n;j++)
                 map[i][j]=(i==j?inf:Get_Dist(i,j));
         double low=,high=40000.0,mid;
         while(high-low>eps){
             mid=(low+high)/;
             if(Judge(mid)){
                 low=mid;
             }else
                 high=mid;
         }
         printf("%.10f\n",mid/);
         for(int i=;i<=n;i++){
             printf(i==n?"%d\n":"%d ",ans[i]);
         }
     }
     return ;
 }