hdu 3622 二分+2-sat

算是比较经典的题目了，精度要适当高一些不然会WA。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

const double eps = 1e-5;
const int N = 200;
const int M = 500000;
double dis[N][N];
int head[N];
int s[N];
bool mark[N];
int n, e, c;

struct Edge
{
    int v, next;
} edge[M];

void addEdge( int u, int v )
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

void init()
{
    e = 0;
    memset( head, -1, sizeof(head) );
}

bool dfs( int u )
{
    if ( mark[u ^ 1] ) return false;
    if ( mark[u] ) return true;
    mark[u] = true;
    s[c++] = u;
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v;
        if ( !dfs(v) ) return false;
    }
    return true;
}

bool solve()
{
    memset( mark, false, sizeof(mark) );
    for ( int i = 0; i < n; i += 2 )
    {
        if ( !mark[i] && !mark[i + 1] )
        {
            c = 0;
            if ( !dfs(i) )
            {
                while ( c )
                {
                    c--;
                    mark[s[c]] = false;
                }
                if ( !dfs( i + 1 ) ) return false;
            }
        }
    }
    return true;
}

struct Point 
{
    int x, y;
} p[N];

double cal( Point a, Point b )
{
    return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );
}

int main ()
{
    while ( scanf("%d", &n) != EOF )
    {
        n = n << 1;
        for ( int i = 0; i < n; i++ )
        {
            scanf("%d%d", &p[i].x, &p[i].y);
        }
        for ( int i = 0; i < n; i++ )
        {
            for ( int j = i + 1; j < n; j++ )
            {
                dis[i][j] = cal( p[i], p[j] );
            }
        }
        double lb = 0, ub = 1e8;
        while ( lb + eps < ub )
        {
            init();
            double mid = ( lb + ub ) / 2.0;
            for ( int i = 0; i < n; i++ )
            {
                for ( int j = i + 1; j < n; j++ )
                {
                    if ( ( i ^ 1 ) == j ) continue;
                    if ( 2.0 * mid > eps + dis[i][j] )
                    {
                        addEdge( i, j ^ 1 );
                        addEdge( j, i ^ 1 );
                    }
                }
            }
            if ( solve() )
            {
                lb = mid;
            }
            else
            {
                ub = mid - eps;
            }
        }
        printf("%.2lf
", lb);
    }
    return 0;
}