poj3241 曼哈顿最小生成树

题目：求平面上n个点的曼哈顿最小生成树的第k大的边。

思路：模板题

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<numeric>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define INF 0x3f3f3f3f
#define mm(a,b) memset(a,b,sizeof(a))
#define PP puts("*********************");
template<class T> T f_abs(T a){ return a > 0 ? a : -a; }
template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
// 0x3f3f3f3f3f3f3f3f
// 0x3f3f3f3f

const int maxn=1e4+50;
const int maxm=2e3+50;
int tree[maxm],pos[maxm];
int lowbit(int i){
    return i&(-i);
}
void add(int i,int val,int id){
    while(i<maxm){
        if(tree[i]>val){
            tree[i]=val;
            pos[i]=id;
        }
        i+=lowbit(i);
    }
}
int query(int i){
    int id=-1,val=INF;
    while(i>0){
        if(tree[i]<val){
            val=tree[i];
            id=pos[i];
        }
        i-=lowbit(i);
    }
    return id;
}
struct Node{
    int id,x,y,xsy;
}point[maxn],p[maxn];
bool cmpx(Node a,Node b){
    if(a.x==b.x) return a.y>b.y;
    return a.x>b.x;
}
bool cmpxsy(Node a,Node b){
    return a.xsy<b.xsy;
}
struct Edge{
    int u,v,w;
    Edge(int _u=0,int _v=0,int _w=0):u(_u),v(_v),w(_w){}
    bool operator<(const Edge& rhs)const{
        return w<rhs.w;
    }
};
vector<Edge> edge;
int dist(Node a,Node b){
    return f_abs(a.x-b.x)+f_abs(a.y-b.y);
}
void Manhaton(int n){
    for(int i=0;i<n;i++)
        point[i].xsy=point[i].x-point[i].y;
    sort(point,point+n,cmpxsy);
    int now=0,pre=-INF;
    for(int i=0;i<n;i++){
        if(pre!=point[i].xsy){
            now++;
            pre=point[i].xsy;
        }
        point[i].xsy=now;
    }
    sort(point,point+n,cmpx);
    for(int i=1;i<maxm;i++)
        tree[i]=INF,pos[i]=-1;
    for(int i=0;i<n;i++){
        int u=point[i].id;
        int v=query(point[i].xsy);
        if(v!=-1) edge.push_back(Edge(u,v,dist(p[u],p[v])));
        add(point[i].xsy,point[i].x+point[i].y,u);
    }
}
void buildEdge(int n){
    edge.clear();
    for(int i=0;i<4;i++){
        for(int j=0;j<n;j++) point[j]=p[j];
        for(int j=0;j<n;j++){
            if(i==1) swap(point[j].x,point[j].y);
            else if(i==2) point[j].y=-point[j].y;
            else if(i==3){
                swap(point[j].x,point[j].y);
                point[j].y=-point[j].y;
            }
        }
        Manhaton(n);
    }
}
int fa[maxn];
int findset(int x){
    if(x==fa[x]) return x;
    return fa[x]=findset(fa[x]);
}
int MST(int n,int k){
    if(n<=k) return 0;
    for(int i=0;i<n;i++) fa[i]=i;
    sort(edge.begin(),edge.end());
    int sz=edge.size();
    for(int i=0;i<sz;i++){
        int u=findset(edge[i].u);
        int v=findset(edge[i].v);
        if(u==v) continue;
        fa[u]=v;
        n--;
        if(n==k) return edge[i].w;
    }
    return 0;
}
int main(){

    int n,k;
    while(~scanf("%d%d",&n,&k)){
        for(int i=0;i<n;i++){
            scanf("%d%d",&p[i].x,&p[i].y);
            p[i].id=i;
        }
        buildEdge(n);
        printf("%d\n",MST(n,k));
    }
    return 0;
}