hdu 3622 Bomb Game【二分+2-SAT+tarjan】

用read()会挂

二分半径，显然最优的是所有原都用这个最小半径，然后2-SAT把相交的圆建图，跑tarjan判一下可行性即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int N=205;
int n,h[N],cnt,dfn[N],low[N],tot,s[N],top,bl[N],col;
bool v[N];
struct qwe
{
	int ne,to;
}e[N*N*8];
struct dian
{
	double x,y;
}a[N],b[N];
double dis(dian a,dian b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int read()
{
	int r=0,f=1;
	char p=getchar();
	while(p>='9'||p<'0')
	{
		if(p=='-')
			f=-1;
		p=getchar();
	}
	while(p>='0'&&p<='9')
	{
		r=r*10+p-48;
		p=getchar();
	}
	return r*f;
}
void add(int u,int v)
{//cerr<<u<<" "<<v<<endl;
	cnt++;
	e[cnt].ne=h[u];
	e[cnt].to=v;
	h[u]=cnt;
}
void tarjan(int u)
{
	dfn[u]=low[u]=++tot;
	v[s[++top]=u]=1;
	for(int i=h[u];i;i=e[i].ne)
	{
		if(!dfn[e[i].to])
		{
			tarjan(e[i].to);
			low[u]=min(low[u],low[e[i].to]);
		}
		else if(v[e[i].to])
			low[u]=min(low[u],dfn[e[i].to]);
	}
	if(low[u]==dfn[u])
	{
		col++;
		while(s[top]!=u)
		{
			bl[s[top]]=col;
			v[s[top--]]=0;
		}
		bl[s[top]]=col;
		v[s[top--]]=0;
	}
}
bool ok(double w)
{//cerr<<w<<endl;
	memset(h,0,sizeof(h));
	memset(v,0,sizeof(v));
	memset(dfn,0,sizeof(dfn));
	cnt=0,tot=0,top=0,col=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
		{
			if(dis(a[i],a[j])<2.0*w)
				add(i,j+n),add(j,i+n);
			if(dis(a[i],b[j])<2.0*w)
				add(i,j),add(j+n,i+n);
			if(dis(b[i],a[j])<2.0*w)
				add(i+n,j+n),add(j,i);
			if(dis(b[i],b[j])<2.0*w)
				add(i+n,j),add(j+n,i);
		}
	for(int i=1;i<=n+n;i++)
		if(!dfn[i])
			tarjan(i);
	for(int i=1;i<=n;i++)
		if(bl[i]==bl[i+n])
			return 0;
	return 1;
}
int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)
			scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&b[i].x,&b[i].y);
		double l=0,r=1e4,ans=0;
		while(r-l>1e-3)
		{
			double mid=(l+r)/2;
			if(ok(mid))
				l=mid,ans=mid;
			else
				r=mid;
		}
		printf("%.2f\n",ans);
	}
	return 0;
}