[POJ 2349] Arctic Network

[题目链接]

        http://poj.org/problem?id=2349

[算法]

         首先，我们发现答案是具有单调性的，所以我们可以二分答案

         检验时，我们将所有距离小于二分值的点连边，然后判断联通块数量是否小于等于S即可

[代码]

         

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h>
using namespace std;
#define MAXP 1010
const int INF = 2e9;
const double eps = 1e-4;

struct edge
{
        int to,nxt;
} e[MAXP * MAXP * 2];

int i,j,S,P,T,tot;
int head[MAXP];
bool visited[MAXP];
pair<double,double> a[MAXP];
double l,r,mid,ans;

inline double dist(pair<double,double> a,pair<double,double> b)
{
        return sqrt((a.first - b.first) * (a.first - b.first) + (a.second - b.second) * (a.second - b.second));        
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot;
}
inline void bfs(int s)
{
        int i,l,r,u,v;
        static int q[MAXP];
        q[l = r = 1] = s;
        while (l <= r)
        {
                u = q[l];
                l++;
                for (i = head[u]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        if (!visited[v]) 
                        {
                                q[++r] = v;
                                visited[v] = true;
                        }
                }
        }
} 
inline bool check(double mid)
{
        int i,j,cnt = 0;
        tot = 0;
        memset(head,0,sizeof(head));
        memset(visited,false,sizeof(visited));
        for (i = 1; i <= P; i++)
        {
                for (j = i + 1; j <= P; j++)
                {
                        if (dist(a[i],a[j]) <= mid)
                        {
                                addedge(i,j);
                                addedge(j,i);        
                        }        
                }        
        }        
        for (i = 1; i <= P; i++)
        {
                if (!visited[i])
                {
                        visited[i] = true;
                        cnt++;
                        bfs(i);
                }
        }
        return cnt <= S; 
}

int main() 
{
        
        scanf("%d",&T);
        while (T--)
        {
                scanf("%d%d",&S,&P);
                for (i = 1; i <= P; i++) scanf("%lf%lf",&a[i].first,&a[i].second);
                l = 1; r = 10000;
                while (r - l > eps)
                {
                        mid = (l + r) / 2;
                        if (check(mid))
                        {
                                ans = mid;
                                r = mid;
                        } else l = mid;
                }
                printf("%.2f\n",ans);
        }
        
        return 0;
    
}