Description
平面坐标系xOy内，给定n个顶点V = (x , y)。对于顶点u、v，u与v之间的距离d定义为|xu – xv| + |yu – yv|
你的任务就是求出这n个顶点的最小生成树。
Input
第一行一个正整数n，表示定点个数。
接下来n行每行两个正整数x、y，描述一个顶点。
Output
只有一行，为最小生成树的边的距离和。
Sample Input
4
1 0
0 1
0 -1
-1 0
Sample Output
6 [数据约定] 对于30%的数据n <= 2000； 对于50%的数据n <= 5000； 对于100%的数据n <= 50000； 0 <= x, y <= 100000。
题解：http://blog.csdn.net/acm_cxlove/article/details/8890003
code：
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #define maxn 50005 7 #define inf 1061109567 8 using namespace std; 9 char ch; 10 bool ok; 11 void read(int &x){ 12 for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1; 13 for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar()); 14 if (ok) x=-x; 15 } 16 int n,m,ans=0; 17 struct Point{ 18 int x,y,d,id; 19 }point[maxn],tmp[maxn]; 20 bool cmp1(Point a,Point b){ 21 if (a.x!=b.x) return a.x>b.x; 22 return a.y>b.y; 23 } 24 int calc(Point a,Point b){return abs(a.x-b.x)+abs(a.y-b.y);} 25 struct Edge{ 26 int u,v,c; 27 }edge[maxn<<2]; 28 bool cmp2(Edge a,Edge b){return a.c<b.c;} 29 int cntd,d[maxn]; 30 struct DATA{ 31 int val,pos; 32 void init(){val=inf,pos=-1;} 33 void update(DATA b){if (val>b.val) val=b.val,pos=b.pos;} 34 }; 35 struct bit{ 36 #define lowbit(x) ((x)&(-(x))) 37 DATA node[maxn]; 38 void init(){for (int i=1;i<=cntd;i++) node[i].init();} 39 void insert(int x,DATA p){ 40 x=cntd-x+1; 41 for (int i=x;i<=cntd;i+=lowbit(i)) node[i].update(p); 42 } 43 int query(int x){ 44 x=cntd-x+1; 45 DATA ans; ans.init(); 46 for (int i=x;i;i-=lowbit(i)) ans.update(node[i]); 47 return ans.pos; 48 } 49 }T; 50 void prepare(){ 51 for (int i=1;i<=n;i++) d[i]=point[i].y-point[i].x; 52 sort(d+1,d+n+1),cntd=unique(d+1,d+n+1)-d-1; 53 for (int i=1;i<=n;i++) point[i].d=lower_bound(d+1,d+cntd+1,point[i].y-point[i].x)-d; 54 sort(point+1,point+n+1,cmp1); 55 T.init(); 56 for (int i=1;i<=n;i++){ 57 int u=point[i].id,v=T.query(point[i].d); 58 if (v!=-1) edge[++m]=(Edge){u,v,calc(tmp[u],tmp[v])}; 59 T.insert(point[i].d,(DATA){point[i].x+point[i].y,u}); 60 } 61 } 62 int fa[maxn]; 63 int find(int x){return x==fa[x]?fa[x]:fa[x]=find(fa[x]);} 64 int main(){ 65 read(n); 66 for (int i=1;i<=n;i++) read(point[i].x),read(point[i].y),point[i].id=i; 67 for (int i=1;i<=n;i++) tmp[i]=point[i]; prepare(); 68 for (int i=1;i<=n;i++) point[i].x=tmp[i].y,point[i].y=tmp[i].x,point[i].id=i; prepare(); 69 for (int i=1;i<=n;i++) point[i].x=-tmp[i].y,point[i].y=tmp[i].x,point[i].id=i; prepare(); 70 for (int i=1;i<=n;i++) point[i].x=tmp[i].x,point[i].y=-tmp[i].y,point[i].id=i; prepare(); 71 sort(edge+1,edge+m+1,cmp2); 72 for (int i=1;i<=n;i++) fa[i]=i; 73 for (int i=1,cnt=0;i<=m&&cnt<n;i++) 74 if (find(edge[i].u)!=find(edge[i].v)) 75 fa[find(edge[i].u)]=find(edge[i].v),cnt++,ans+=edge[i].c; 76 printf("%d ",ans); 77 return 0; 78 }