ABC065D Built[最小生成树]

这题和某道最短路题神似。对于任意点对，将他们连边，不如将他们分别沿$x,y$轴方向上点按顺序连起来，这样不仅可能多连通一些点，也花费更低，所以按照最短路那题的连边方式跑一个kruskal就行了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #define dbg(x) cerr << #x << " = " << x <<endl
 using namespace std;
 typedef long long ll;
 typedef double db;
 typedef pair<int,int> pii;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,):;}
 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}
 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const int N=+;
 struct thxorz{
     int u,v,w;
     thxorz(int u=,int v=,int w=):u(u),v(v),w(w){}
     inline bool operator <(const thxorz&A)const{return w<A.w;}
 }e[N<<];
 int n,tot;
 struct stothx{int x,y,id;}A[N];
 inline bool cmp1(stothx a,stothx b){return a.x<b.x;}
 inline bool cmp2(stothx a,stothx b){return a.y<b.y;}
 int anc[N];
 ll ans;
 inline int get_anc(int x){return anc[x]==x?x:anc[x]=get_anc(anc[x]);}

 int main(){//freopen("test.in","r",stdin);//freopen("test.ans","w",stdout);
     read(n);
     for(register int i=;i<=n;++i)read(A[i].x),read(A[i].y),A[i].id=i;
     sort(A+,A+n+,cmp1);
     for(register int i=;i<n;++i)e[++tot]=thxorz(A[i].id,A[i+].id,A[i+].x-A[i].x);
     sort(A+,A+n+,cmp2);
     for(register int i=;i<n;++i)e[++tot]=thxorz(A[i].id,A[i+].id,A[i+].y-A[i].y);
     sort(e+,e+tot+);
     for(register int i=;i<=n;++i)anc[i]=i;
     for(register int i=;i<=tot;++i)if(get_anc(e[i].u)^get_anc(e[i].v))ans+=e[i].w,anc[anc[e[i].u]]=anc[e[i].v];
     printf("%lld\n",ans);
     return ;
 }