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,0,sizeof(color));
    queue<int>que;
    for(int i=1;i<=n;i++){
        if(color[i]==0){
            color[i]=1;
            que.push(i);
            while(!que.empty()){
                int u=que.front();
                que.pop();
                for(int v=1;v<=n;v++){
                    if(map[u][v]>limit-eps)continue;
                    if(color[u]==color[v])return false;
                    if(color[v]==0){
                        color[v]=3-color[u];
                        que.push(v);
                    }
                }
            }
        }
    }
    for(int i=1;i<=n;i++)ans[i]=color[i];
    return true;
}

    

int main()
{
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                map[i][j]=(i==j?inf:Get_Dist(i,j));
        double low=0,high=40000.0,mid;
        while(high-low>eps){
            mid=(low+high)/2;
            if(Judge(mid)){
                low=mid;
            }else
                high=mid;
        }
        printf("%.10f
",mid/2);
        for(int i=1;i<=n;i++){
            printf(i==n?"%d
":"%d ",ans[i]);
        }
    }
    return 0;
}