老oj曼哈顿最小生成树

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：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#define maxn 50005
#define inf 1061109567
using namespace std;
char ch;
bool ok;
void read(int &x){
    for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;
    for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
    if (ok) x=-x;
}
int n,m,ans=0;
struct Point{
    int x,y,d,id;
}point[maxn],tmp[maxn];
bool cmp1(Point a,Point b){
    if (a.x!=b.x) return a.x>b.x;
    return a.y>b.y;
}
int calc(Point a,Point b){return abs(a.x-b.x)+abs(a.y-b.y);}
struct Edge{
    int u,v,c;
}edge[maxn<<2];
bool cmp2(Edge a,Edge b){return a.c<b.c;}
int cntd,d[maxn];
struct DATA{
    int val,pos;
    void init(){val=inf,pos=-1;}
    void update(DATA b){if (val>b.val) val=b.val,pos=b.pos;}
};
struct bit{
    #define lowbit(x) ((x)&(-(x)))
    DATA node[maxn];
    void init(){for (int i=1;i<=cntd;i++) node[i].init();}
    void insert(int x,DATA p){
        x=cntd-x+1;
        for (int i=x;i<=cntd;i+=lowbit(i)) node[i].update(p);
    }
    int query(int x){
        x=cntd-x+1;
        DATA ans; ans.init();
        for (int i=x;i;i-=lowbit(i)) ans.update(node[i]);
        return ans.pos;
    }
}T;
void prepare(){
    for (int i=1;i<=n;i++) d[i]=point[i].y-point[i].x;
    sort(d+1,d+n+1),cntd=unique(d+1,d+n+1)-d-1;
    for (int i=1;i<=n;i++) point[i].d=lower_bound(d+1,d+cntd+1,point[i].y-point[i].x)-d;
    sort(point+1,point+n+1,cmp1);
    T.init();
    for (int i=1;i<=n;i++){
        int u=point[i].id,v=T.query(point[i].d);
        if (v!=-1) edge[++m]=(Edge){u,v,calc(tmp[u],tmp[v])};
        T.insert(point[i].d,(DATA){point[i].x+point[i].y,u});
    }
}
int fa[maxn];
int find(int x){return x==fa[x]?fa[x]:fa[x]=find(fa[x]);}
int main(){
    read(n);
    for (int i=1;i<=n;i++) read(point[i].x),read(point[i].y),point[i].id=i;
    for (int i=1;i<=n;i++) tmp[i]=point[i]; prepare();
    for (int i=1;i<=n;i++) point[i].x=tmp[i].y,point[i].y=tmp[i].x,point[i].id=i; prepare();
    for (int i=1;i<=n;i++) point[i].x=-tmp[i].y,point[i].y=tmp[i].x,point[i].id=i; prepare();
    for (int i=1;i<=n;i++) point[i].x=tmp[i].x,point[i].y=-tmp[i].y,point[i].id=i; prepare();
    sort(edge+1,edge+m+1,cmp2);
    for (int i=1;i<=n;i++) fa[i]=i;
    for (int i=1,cnt=0;i<=m&&cnt<n;i++)
        if (find(edge[i].u)!=find(edge[i].v)) 
            fa[find(edge[i].u)]=find(edge[i].v),cnt++,ans+=edge[i].c;
    printf("%d
",ans);
    return 0;
}