UVA1146 Now or later

二分枚举答案，2-SAT检验是否可行。

#include 
using namespace std;
const int N=2e3+5;
int n,l,r,mid,ans,tot,a[N*2];
struct number{
     int x,y;}num[N];
int now,top,col,dfn[N*2],low[N*2],sta[N*2],color[N*2];
int cnt,head[N*2];
struct edge{
     int next,to;}e[N*N*4]; 

inline void add(int u,int v)
{
     
	cnt++;
	e[cnt].next=head[u];
	e[cnt].to=v;
	head[u]=cnt;
}

void tarjan(int u)
{
     
	dfn[u]=low[u]=++now;
	sta[++top]=u;
	for (register int i=head[u]; i; i=e[i].next)
	{
     
		if (!dfn[e[i].to])
		{
     
			tarjan(e[i].to);
			low[u]=min(low[u],low[e[i].to]);
		}
		else if (!color[e[i].to]) low[u]=min(low[u],low[e[i].to]);
	}
	if (dfn[u]==low[u])
	{
     
		col++;
		while (sta[top]!=u) color[sta[top]]=col,top--;
		color[sta[top]]=col,top--;
	}
}

inline bool jay(int mid)
{
     
	cnt=0; for (register int i=1; i<=2*n; ++i) head[i]=0;
	for (register int i=1; i<=n; ++i)
	{
     
		for (register int j=1; j<=n; ++j)
		if (i!=j)
		{
     
			if (abs(num[i].x-num[j].x)<mid) add(i,j+n); 
			if (abs(num[i].x-num[j].y)<mid) add(i,j);
			if (abs(num[i].y-num[j].x)<mid) add(i+n,j+n);
			if (abs(num[i].y-num[j].y)<mid) add(i+n,j);
		}
	}
	now=top=col=0;
	for (register int i=1; i<=2*n; ++i) dfn[i]=low[i]=color[i]=0;
	for (register int i=1; i<=2*n; ++i) if (!dfn[i]) tarjan(i);
	for (register int i=1; i<=n; ++i) if (color[i]==color[i+n]) return false;
	return true;
}

int main(){
     
	while (~scanf("%d",&n))
	{
     
		for (register int i=1; i<=n; ++i) scanf("%d%d",&num[i].x,&num[i].y);
		l=0; r=1e7; ans=0;
		while (l<=r)
		{
     
			mid=l+r>>1;
			if (jay(mid)) ans=mid,l=mid+1;
			else r=mid-1;
		}
		printf("%d\n",ans);
	}
return 0;
}