CSA Round #84 The Sprawl

Analysis

曼哈顿距离（$L1$ metric）最小生成树。
Implementation

下面的代码参考了 gispzjz 在比赛中的提交。
#include <bits/stdc++.h>

using namespace std;

#define pb push_back

#define eb emplace_back

#define all(x) x.begin(), x.end()

#define debug(x) cerr << #x <<": " << (x) << endl

#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)

#ifdef LOCAL

#define see(x) cout  << #x << ": " << (x) << endl

#endif

#ifndef LOCAL

#define see(x)

#endif

#define rep(n) for(int _ = 0; _ != (n); ++_)

//#define rep(i, a, b) for(int i = (a); i <= (b); ++i)

#define Rng(i, n) for(int i = 0; i != (n); ++i)

#define rng(i, a, b) for(int i = (a); i < (b); ++i)

#define rno(i, b) for(int i = 0; i<(b); ++i)

#define rnc(i, a, b) for(int i = (a); i<=(b); ++i)

#define RNG(i, a) for(auto &i: (a))

#define dwn(i, r, l) for(int i = (r); i>=(l); i--)

namespace std {

    template<class T>

    T begin(std::pair<T, T> p)

    {

        return p.first;

    }

    template<class T>

    T end(std::pair<T, T> p)

    {

        return p.second;

    }

}

#if __cplusplus < 201402L

template<class Iterator>

std::reverse_iterator<Iterator> make_reverse_iterator(Iterator it)

{

    return std::reverse_iterator<Iterator>(it);

}

#endif

template<class Range>

std::pair<std::reverse_iterator<decltype(begin(std::declval<Range>()))>, std::reverse_iterator<decltype(begin(std::declval<Range>()))>> make_reverse_range(Range &&r)

{

    return std::make_pair(make_reverse_iterator(::begin(r)), make_reverse_iterator(::end(r)));

}

#define RRNG(x, cont) for (auto &x: make_reverse_range(cont))

template<class T> int sign(const T &a) { return a == 0 ? 0 : a > 0 ? 1 : -1; }

template<class T> inline T min(T a, T b, T c){return min(min(a, b), c);}

template<class T> inline T max(T a, T b, T c){return max(max(a, b), c);}

template<class T> void Min(T &a, const T &b){ a = min(a, b); }

template<class T> void Max(T &a, const T &b){ a = max(a, b); }

template<typename T> void println(const T &t) { cout << t << '\n'; }

template<typename T, typename ...Args> void println(const T &t, const Args &...rest) { cout << t << ' '; println(rest...); }

template<typename T> void print(const T &t) { cout << t << ' '; }

template<typename T, typename ...Args> void print(const T &t, const Args &...rest) { cout << t; print(rest...); }

// this overload is chosen when there's only one argument

template<class T> void scan(T &t) { cin >> t; }

template<class T, class ...Args> void scan(T &a, Args &...rest) { cin >> a; scan(rest...); }

using ll = long long;

using ull = unsigned long long;

using vec = vector<ll>;

using mat = vector<vec>;

using pii = pair<int, int>;

using pdd = pair<double, double>;

using pip = pair<int, pii>;

using szt = size_t;

using vi = vector<int>;

using vl = vector<ll>;

using vb = vector<bool>;

using vpii = vector<pii>;

using vvi = vector<vi>;

using pli = pair<ll,int>;

using wg = vector<vpii>; //weighted graph

int cas;

const double pi = acos(-1);

ll mod = 1e9 + 7;

//要求：0<=a<mod, 0<=b<=mod

template<class T>

inline void add_mod(T &a, const T &b) {

    a += b;

    if (a >= mod) a -= mod;

}

template<class T>

void sub_mod(T &a, const T &b){

    a -= b;

    if (a < 0) a += mod;

}

auto bo=[](int x){

    bitset<5> a(x);

    cout << a << endl;

};

//返回值：a中比k小的元素有多少个？

template<class V, class Cont>

int get_rank(const V &k, const Cont &a){

    return std::lower_bound(all(a), k) - a.begin();

}

mat operator*(const mat &a, const mat &b) {

    mat c(a.size(), vec(b[0].size()));

    for (size_t i = 0; i < a.size(); i++) {

        for (size_t j = 0; j < a[0].size(); j++) {

            if (a[i][j]) { // optimization for sparse matrix

                for (size_t k = 0; k < b[0].size(); k++) {

                    add_mod(c[i][k], a[i][j] * b[j][k] % mod);

                }

            }

        }

    }

    return c;

}

vec operator*(const mat &a, const vec &b) {

    vec c(a.size());

    for (size_t i = 0; i < a.size(); i++) {

        for (size_t j = 0; j < a[0].size(); j++) {

            add_mod(c[i], a[i][j] * b[j] % mod);

        }

    }

    return c;

}

mat pow(mat a, ull n) {

    mat res(a.size(), vec(a[0].size()));

    for (size_t i = 0; i < a.size(); i++) {

        res[i][i] = 1;

    }

    while (n) {

        if (n & 1) {

            res = res * a;

        }

        a = a * a;

        n >>= 1;

    }

    return res;

}

// Codeforces does not support __int128

//std::ostream& operator<<(std::ostream& os, __int128 T) {

//    if (T<0) os<<"-";

//    if (T>=10 ) os<<T/10;

//    if (T<=-10) os<<(-(T/10));

//    return os<<( (int) (T%10) >0 ? (int) (T%10) : -(int) (T%10) ) ;

//}

//

//__int128 LPOW(__int128 x, ll n) {

//    __int128 res = 1;

//    for (; n; n /= 2, x *= x, x %= mod) {

//        if (n & 1) {

//            res *= x;

//            res %= mod;

//        }

//    }

//    return res;

//}

ll POW(ll x, ll n){

    ll res = 1;

    for (; n; n /= 2, x *= x, x %= mod) {

        if (n & 1) {

            res *= x;

            res %= mod;

        }

    }

    return res;

}

ll INV(ll x) {

    return POW(x, mod - 2);

}

ll inv(ll x){

//    see(x);

    return x == 1? 1: (mod - mod/x * inv(mod%x) % mod);

}

// 2D rotation

void rotate(double &x, double &y, double theta) {

    double tx = cos(theta) * x - sin(theta) * y;

    double ty = sin(theta) * x + cos(theta) * y;

    x = tx, y = ty;

}

struct dsu{

    vector<int> par;

    explicit dsu(int n){ // 0-indexed

        par.resize(n);

        rng(i, 0, n){

            par[i] = i;

        }

    }

    bool same(int x, int y){

        return root(x) == root(y);

    }

    int root(int x){

        return par[x] == x ? x : par[x] = root(par[x]);

    }

    void unite(int x, int y){

        x = root(x);

        y = root(y);

        par[x] = y;

    }

};

struct bit {

    vector<int> a;

    vector<int> id; // id[i]：键区间(i-lowbit(i), i]中，x+y取最小值的点的编号

    bit(int n, int v = 0, bool manhattan = false) {

        a.resize(n + 1);

        for (int i = 1; i <= n; ++i) a[i] = v;

        if(manhattan){

            id.resize(n+1);

            rng(i, 1, n+1) id[i] = -1;

        }

    }

    ll sum(int x) {

        ll res = 0;

        while (x) {

            res += a[x];

            x -= x & -x;

        }

        return res;

    }

    ll sum(int l, int r) {

        if (l > r) return 0;

        return sum(r) - sum(l - 1);

    }

    void add(int x, ll v) {

        while (x < a.size()) {

            a[x] += v;

            x += x & -x;

        }

    }

    void min(int x, int v) {

        while (x < a.size()) {

            a[x] = std::min(a[x], v);

            x += x & -x;

        }

    }

    int manhattan_min(int x){ // 返回x+y最小的点的编号

        int ans = -1;

        int res = INT_MAX;

        while(x){

            if(a[x] < res){

                res = a[x];

                ans = id[x];

            }

            x -= x & - x;

        }

        return ans;

    }

    void manhattan_min(int x, int v, int i){

        while(x < a.size()){

            if(a[x] > v){

                a[x] = v;

                id[x] = i;

            }

            x += x & - x;

        }

    }

    void max(int x, int v) {

        while (x < a.size()) {

            a[x] = std::max(a[x], v);

            x += x & -x;

        }

    }

    int min(int x) {

        int res = INT_MAX;

        while (x) {

            res = std::min(res, a[x]);

            x -= x & -x;

        }

        return res;

    }

    int max(int x) {

        int res = INT_MIN;

        while (x) {

            res = std::max(res, a[x]);

            x -= x & -x;

        }

        return res;

    }

};

namespace util{

    int len(ll x){return snprintf(nullptr, 0, "%lld", x);}

    vi get_d(ll x){

        vi res;

        while(x) {

            res.pb(x%10);

            x /= 10;

        }

        reverse(all(res));

        return res;

    }

    template <class T> T parity(const T &a){

        return a & 1;

    }

    template <class T>

    void out (const vector<T> &a){

        std::copy(a.begin(), a.end(), std::ostream_iterator<T>(std::cout, ", "));

        cout << endl;

    };

}

using namespace util;

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using order_statistic_tree = __gnu_pbds::tree<

        int,

        __gnu_pbds::null_type,

        greater<int>,

        __gnu_pbds::rb_tree_tag,

        __gnu_pbds::tree_order_statistics_node_update>;

const ll LINF = LLONG_MAX/10;

const int INF = INT_MAX/10;

const int M = 5005;

const int N = 1e5+5;

int a[N];

struct point{

    int x, y, id;

    bool operator<(const point &rhs)const{

        return x < rhs.x || (x == rhs.x && y < rhs.y);

    }

    int dis(const point rhs) {

        return abs(x - rhs.x) + abs(y - rhs.y);

    };

};

struct edge{

    int u, v, l;

    bool operator<(const edge &rhs) {

        return l < rhs.l;

    };

};

ll kruskal(vector<edge> &e, int n) { // 0-indexed

    dsu a(n);

    vi size(n);

    fill(all(size), 1);

    sort(all(e));

    ll sum = 0;

    int cnt = 0;

    RNG(x, e) {

        int u = a.root(x.u), v = a.root(x.v);

        if(u != v){

            sum += 1LL * x.l/2 * size[u] * size[v];

            a.par[u] = v;

            size[v] += size[u];

            ++cnt;

            if(cnt == n - 1) break;

        }

    }

    return sum;

};

void R1(vector<point>& a, vector<edge> &e){

    // 离散化

    map<int, int> ls;

    RNG(p, a) {

        ls[p.x - p.y];

    }

    int cnt = 0;

    RNG(x, ls) {

        x.second = ++cnt;

    }

    bit b(cnt, INT_MAX, true);

    sort(all(a));

    dwn(i, a.size() - 1, 0) {

        // 为了好写，规定区域 R_1(s) 为 x >= x_s, x - y <= x_s - y_s

        int pos = ls[a[i].x- a[i].y];

        int j = b.manhattan_min(pos);

        if (j != -1) {

            e.pb({a[i].id, a[j].id, a[i].dis(a[j])});

        }

        b.manhattan_min(pos, a[i].x + a[i].y, i);

    }

}

ll manhattan(vector<point> &a){

   vector<edge> e;

   rng(i, 0, 4){

       if(i == 1 || i == 3){ // 交换x,y坐标

           RNG(j, a){

               swap(j.x, j.y);

           }

       }

       else if(i == 2){

           RNG(j, a){

               j.x = - j.x;

           }

       }

       R1(a, e);

   }

   return kruskal(e, a.size());

}

int main() {

    // Single Cut of Failure taught me

    cout << std::fixed;

    cout << setprecision(10);

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

#ifdef LOCAL

    freopen("main.in", "r", stdin);

//    freopen("main.out", "w", stdout);

#endif

    int n;

    scan(n);

    vector<point> p(n);

    rng(i, 0, n){

        scan(p[i].x, p[i].y);

        p[i].id = i;

    }

    println(manhattan(p));

#ifdef LOCAL

    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

#endif

    return 0;

}
备忘

$R_1(s)\colon x \ge x_s, x - y \le x_s - y_s$，

$R_2(s)\colon y \ge y_s, x - y \ge x_s - y_s$，

$R_3(s) \colon y \le y_s, x + y \ge x_s + y_s$，

$R_4(s) \colon x \ge x_s, x+y \le x_s + y_s $ 。
$R_2$ 变换到 $R_1$： $(x,y) \to (y, x)$

$R_3$ 变换的 $R_1$： $(x, y) \to (-y, x)$

$R_4$ 变换到 $R_1$：$(x, y) \to (x, -y)$
$R_1, R_2$ 的离散化部分可共用， $R_3, R_4$ 的离散化部分可共用，但上面给出的实现并未这样做，这是一个可以优化的点。
扩展

欧几里得距离最小生成树也可用类似的划分平面的方法解决。

仍然可以按 $45^\circ$ 度划分平面。$\forall a, b \in R_1(s)$，在 $\triangle sab$ 中，有 $\angle{s} < \max(\angle{a}, \angle{b})$，于是由正弦定理可知 $|ab| < \max(|sa|,|sb|)$ 。
巴特西

CSA Round #84 The Sprawl

Analysis

Implementation

备忘

扩展

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