曼哈顿最小生成树 全网最全

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFFLL;  //2147483647
const ll nmos = 0x80000000LL;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18
const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/

const int maxn = 1e5+9;
struct Point{
    int x,y,id;
    
    bool operator < (const Point &q)const{
        if(x == q.x)return y < q.y;
        return x < q.x;
    }
}p[maxn];

struct BITnode{
    int w,p;
}BITsum[maxn];

struct Edge{
    int u,v,dis;//曼哈顿距离
    bool operator < (const Edge & e)const {
        return dis < e.dis;
    }
} e[maxn << 3 | 1];

int tot = 0,fa[maxn];
int find(int x){
    if(fa[x] == x)return x;
    else return fa[x] = find(fa[x]);
}
void addedge(Point a,Point b){
    tot ++;
    e[tot].u = a.id;
    e[tot].v = b.id;
    e[tot].dis = abs(a.x - b.x) + abs(a.y-b.y);
}
int lowbit(int x){
    return x & (-x);
}
int a[maxn],n,k,*ix[maxn];
bool cmp(int * x,int * y){
    return (*x) < (*y);
}
int query(int x){
    int mx = inf,p = -1;
    while(x <= n){
        if(BITsum[x].w < mx){
            mx = BITsum[x].w;
            p = BITsum[x].p;
        }
        x += lowbit(x);
    }
    return p;
}

void update(int x,int w,int p){
    while(x > 0){
        if(BITsum[x].w > w){
            BITsum[x].w = w;
            BITsum[x].p = p;
        }
        x -= lowbit(x);
    }
}

int main(){
    
    scanf("%d%d", &n, &k);
    
    for(int i = 1; i <= n; i++){
        scanf("%d%d", &p[i].x, &p[i].y);
        p[i].id = i;
    }
    // 建立边
    tot = 0;
    for(int dir = 1; dir <= 4; dir++){
        
        if(dir==2 || dir==4)
            for(int i=1; i<=n; i++)swap(p[i].x,p[i].y);
        else if(dir == 3)
            for(int i=1; i<=n; i++)p[i].x = -p[i].x;
        
        sort(p + 1, p + n + 1);
        //这里离散的操作我觉得也很厉害。
        for(int i=1; i<=n; i++)a[i] = p[i].y - p[i].x, ix[i] = &a[i];
        sort(ix + 1, ix + n + 1, cmp);
        
        for(int i=1; i<=n; i++) *ix[i] = i;
        for(int i = 1; i <= n; i++)
            BITsum[i].w = inf,BITsum[i].p = -1;
        
        for(int i = n; i >= 1; i--){
            int pos = query(a[i]);
            if(pos!=-1){
                addedge(p[i], p[pos]);
            }
            update(a[i], p[i].x + p[i].y, i);
        }
        
    }
    //库鲁斯卡尔 算法 开始
    sort(e + 1,e + tot + 1);
    
    for(int i=1; i<=n; i++)fa[i] = i;
    
    int cnt = n - k,c1 = 0,ans = 0;
    for(int i=1; i<= tot; i++){
        if( find(e[i].u) != find(e[i].v) ){
            int px = find(e[i].u);
            int py = find(e[i].v);
            fa[px] = py;
            c1++;
            if(c1 == cnt){ ans = e[i].dis;break;}
        }
    }
    printf("%d\n", ans);
    
    return 0;
}