CSA Round #84 The Sprawl

题目

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
",__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 << '
'; }
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.
";
#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 $R_1$： $(x,y) o (y, x)$
$R_3$ 变换的 $R_1$： $(x, y) o (-y, x)$
$R_4$ 变换到 $R_1$：$(x, y) o (x, -y)$

$R_1, R_2$ 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 $ riangle sab$ 中，有 $angle{s} < max(angle{a}, angle{b})$，于是由正弦定理可知 $|ab| < max(|sa|,|sb|)$ 。