题意：
给n个村庄的坐标和两个特殊点s1,s2的坐标，现在要将每个村庄连到s1或s2上，使n个村庄间的最大距离最小。
分析：
二分任意两村庄间的最大距离，用2-sat判断该最大距离是否可行，二分的时候可以顺便记录答案，不用等最后区间为空时再输出l或l-1。
代码：
//poj 2749
//sep9
#include <iostream>
#include <vector>
using namespace std;
const int maxN=1024;
const int maxL=4000000;
vector<int> g[maxN],ng[maxN];
int n,m,cnt,scc,vis[maxN],dfn[maxN];
int x1,y1,x2,y2;
int x[maxN],y[maxN];
int a,b;
int ap[maxN],aq[maxN],bp[maxN],bq[maxN];
int d[maxN],D;
void addegde(int u,int v)
{g[u].push_back(v);ng[v].push_back(u);
}void dfs(int k)
{vis[k]=1;for(int i=g[k].size()-1;i>=0;--i)if(!vis[g[k][i]])dfs(g[k][i]);dfn[++cnt]=k;
}void ndfs(int k)
{vis[k]=scc;for(int i=ng[k].size()-1;i>=0;--i)if(!vis[ng[k][i]])ndfs(ng[k][i]);
}
void kosaraju()
{memset(vis,0,sizeof(vis));cnt=0;for(int i=1;i<=2*n;++i)if(!vis[i])dfs(i); memset(vis,0,sizeof(vis));scc=0;for(int i=2*n;i>=1;--i)if(!vis[dfn[i]]){++scc;ndfs(dfn[i]);}
}int two_sat(int dis)
{int i,j;for(i=1;i<=2*n;++i)g[i].clear(),ng[i].clear();for(i=1;i<=a;++i){int p=ap[i],q=aq[i];addegde(p,q+n);addegde(p+n,q);addegde(q,p+n);addegde(q+n,p); }for(i=1;i<=b;++i){int p=bp[i],q=bq[i];addegde(p,q);addegde(q,p);addegde(p+n,q+n);addegde(q+n,p+n);}for(i=1;i<=n;++i)for(j=i+1;j<=n;++j)if(i!=j){if(d[i]+d[j]>dis){addegde(i,j+n);addegde(j,i+n);} if(d[i]+d[j+n]+D>dis){addegde(j+n,i+n);addegde(i,j);} if(d[i+n]+d[j]+D>dis){addegde(i+n,j+n);addegde(j,i);}if(d[i+n]+d[j+n]>dis){addegde(i+n,j);addegde(j+n,i);} }kosaraju();for(i=1;i<=n;++i)if(vis[i]==vis[i+n])return 0; return 1;
}int main()
{int i,j;scanf("%d%d%d",&n,&a,&b);scanf("%d%d%d%d",&x1,&y1,&x2,&y2);for(i=1;i<=n;++i)scanf("%d%d",&x[i],&y[i]);for(i=1;i<=n;++i){d[i]=abs(x[i]-x1)+abs(y[i]-y1);d[i+n]=abs(x[i]-x2)+abs(y[i]-y2);}D=abs(x1-x2)+abs(y1-y2);int p,q;for(i=1;i<=a;++i)scanf("%d%d",&ap[i],&aq[i]); for(i=1;i<=b;++i)scanf("%d%d",&bp[i],&bq[i]);int l=0,r=maxL,ans=-1; while(l<r){int mid=(l+r)/2;if(two_sat(mid)==0)l=mid+1;elser=mid,ans=mid;}printf("%d",ans);return 0;
}