bzoj 1052: [HAOI2007]覆盖问题 & Luogu P2218 [HAOI2007]覆盖问题 题解

#include<algorithm>
#include<bitset>
#include<complex>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<iterator>
#include<limits>
#include<list>
#include<locale>
#include<map>
#include<memory>
#include<new>
#include<numeric>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<typeinfo>
#include<utility>
#include<valarray>
#include<vector>
#include<cctype>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<climits>
#include<clocale>
#include<cmath>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
using namespace std;

#define re register
#define It std::set<int>::iterator
//#define int long long 

class Quick_Input_Output{
    private:
        static const int S=1<<21;
    //  #define gc() (A==B&&(B=(A=Rd)+fread(Rd,1,S,stdin),A==B)?EOF:*A++)
        char Rd[S],*A,*B;
        #define pc putchar
    public:
    //  #undef gc()
        #define gc getchar 
        inline int read(){
            int res=0,f=1;char ch=gc();
            while(ch<'0'ch>'9'){if(ch=='-') f=-1;ch=gc();}
            while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=gc();
            return res*f;
        }
        inline void write(int x){
            if(x<0) pc('-'),x=-x;
            if(x<10) pc(x+'0');
            else write(x/10),pc(x%10+'0');
        }
        #undef gc
        #undef pc
}I;

int n,vis[20010];
struct node{int x,y;}a[20010];
inline bool cmp(node qx,node qy){return qx.x<qy.x(qx.x==qy.x&&qx.y<qy.y);}
inline bool dfs(int x,int L){
    re int Minx=2e9,Maxx=-2e9,Miny=2e9,Maxy=-2e9;//求出大正方形的四个点的坐标
    for(int i=1;i<=n;i++){
        if(vis[i]==0){
            Maxx=max(Maxx,a[i].x);
            Minx=min(Minx,a[i].x);
            Maxy=max(Maxy,a[i].y);
            Miny=min(Miny,a[i].y);
        }
    }
    re int lenx=Maxx-Minx,leny=Maxy-Miny;//如果大正方形的边长<=L那么肯定可以
    if(max(lenx,leny)<=L) return 1;
    if(x==3) return 0;//如果已经放了3个正方形却不行，那么就肯定不行
    for(int i=1;i<=n;i++){if(vis[i]==0&&Minx<=a[i].x&&a[i].x<=Minx+L&&Miny<=a[i].y&&a[i].y<=Miny+L) vis[i]=x;}//4个角的情况，分别dfs
    if(dfs(x+1,L)) return 1;
    for(int i=1;i<=n;i++){if(vis[i]==x) vis[i]=0;}
    for(int i=1;i<=n;i++){if(vis[i]==0&&Maxx-L<=a[i].x&&a[i].x<=Maxx&&Miny<=a[i].y&&a[i].y<=Miny+L) vis[i]=x;}
    if(dfs(x+1,L)) return 1;
    for(int i=1;i<=n;i++){if(vis[i]==x) vis[i]=0;}
    for(int i=1;i<=n;i++){if(vis[i]==0&&Minx<=a[i].x&&a[i].x<=Minx+L&&Maxy-L<=a[i].y&&a[i].y<=Maxy) vis[i]=x;}
    if(dfs(x+1,L)) return 1;
    for(int i=1;i<=n;i++){if(vis[i]==x) vis[i]=0;}
    for(int i=1;i<=n;i++){if(vis[i]==0&&Maxx-L<=a[i].x&&a[i].x<=Maxx&&Maxy-L<=a[i].y&&a[i].y<=Maxy) vis[i]=x;}
    if(dfs(x+1,L)) return 1;
    for(int i=1;i<=n;i++){if(vis[i]==x) vis[i]=0;}
    return 0;
}
inline bool check(int x){memset(vis,0,sizeof(vis));return dfs(1,x);}//每次check前清空
signed main(){
    n=I.read();
    for(int i=1;i<=n;i++) a[i].x=I.read(),a[i].y=I.read();
    sort(a+1,a+n+1,cmp);
    int l=0,r=2e9,ans=2e9;//二分
    while(l<=r){
        re int mid=l+r>>1;
        if(check(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    I.write(ans);putchar('\n');
    return 0;
}