ZOJ 3717

这题是二分+2SAT.

总结一下SAT题的特征。首先，可能会存在二选一的情况，然后会给出一些矛盾。据这些矛盾加边，再用SAT判定。

这一道题好像不能直接用printf("%0.3lf")，因为这个是四舍五入的，这道题好像不能四舍五入，只好选减去0.0005再按这个格式输出了。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <stdlib.h>
 #define oo 1000000007
 #define eps 1e-4
 using namespace std;
 const int MAXN=;
 const int MAXM=;
 struct b{
     int x,y,z;
 }ball[MAXN];
 double dis[MAXN][MAXN];
 int head[MAXN],dfn[MAXN],low[MAXN],tot,stop,indx,belong[MAXN],pat;
 int st[MAXN];
 bool stack[MAXN];
 int n;
 struct e{
     int u,v;
     int next;
 }edge[MAXM];
 
 void addedge(int u,int v){
     edge[tot].u=u;
     edge[tot].v=v;
     edge[tot].next=head[u];
     head[u]=tot++;
 }
 
 void tarjan(int u){
     int v;
     dfn[u]=low[u]=++indx;
     st[stop++]=u;
     stack[u]=true;
     for (int e=head[u];e!=-;e=edge[e].next){
         v=edge[e].v;
         if (dfn[v]==) {
             tarjan(v) ;
             low[u] = min(low[u], low[v]) ;
         }
         else if (stack[v]) {
             low[u] = min(low[u], dfn[v]) ;
         }
     }
     if (dfn[u] == low[u]) {
         pat++;
         do{
             v = st[--stop];
             belong[v]=pat;
             stack[v]=false;
         }while(u!= v);
     }
 }
 
 bool slove(double leng){
     int a,b; tot=; indx=; stop=; pat=-;
     memset(head,-,sizeof(head));
     for(int i=;i<n;i++){
         a=i;
         for(int j=i+;j<n;j++){
             b=j;
             if(leng*>dis[a*][b*]){
                 addedge(*a+,b*);
                 addedge(b*+,a*);
             }
             if(leng*>dis[a*][b*+]){
                 addedge(a*+,b*+);
                 addedge(b*,a*);
             }
             if(leng*>dis[a*+][b*]){
                 addedge(a*,b*);
                 addedge(b*+,a*+);
             }
             if(leng*>dis[a*+][b*+]){
                 addedge(a*,b*+);
                 addedge(b*,a*+);
             }
         }
     }
     memset(dfn,,sizeof(dfn));
     memset(low,,sizeof(low));
     memset(stack,false,sizeof(stack));
     memset(belong,-,sizeof(belong));
     for(int i=;i<*n;i++){
         if(dfn[i]==){
             tarjan(i);
         }
     }
     bool flag=true;
     for(int i=;i<n;i++){
         if(belong[i*]==belong[i*+]){
             flag=false;
             break;
         }
     }
     return flag;
 }
 
 int main(){
     double xi,yi,zi;
     while(scanf("%d",&n)!=EOF){
         for(int i=;i<*n;i++){
             scanf("%d%d%d",&ball[i].x,&ball[i].y,&ball[i].z);
             i++;
             scanf("%d%d%d",&ball[i].x,&ball[i].y,&ball[i].z);
         }
         double high=oo,lown=; double tmp;
         for(int i=;i<*n;i++){
             for(int j=i;j<*n;j++){
                 xi=ball[i].x-ball[j].x;
                 yi=ball[i].y-ball[j].y;
                 zi=ball[i].z-ball[j].z;
                 tmp=sqrt(xi*xi+yi*yi+zi*zi);
                 dis[i][j]=dis[j][i]=tmp;
             }
         }
         double ans;
         while(lown<high-eps){
             double mid=(high+lown)/;
             if(slove(mid)){
                 ans=mid;
                 lown=mid;
             }
             else high=mid;
         }
         printf("%0.3lf\n",ans-0.0005);
     }
     return ;
 }