#1007840 (D) No.2873 Kendall's Tau

• 問題: No.2873 Kendall's Tau
• ユーザー: InTheBloom
• 提出日時: 2024-09-06 22:51:33
• 言語: D (dmd 2.109.1)
• 結果: AC
• 実行時間: 845 ms / 4,500 ms
• コード長: 7,687 bytes
• コンパイル時間: 6,288 ms
• コンパイル使用メモリ: 234,592 KB
• 実行使用メモリ: 19,884 KB
• 最終ジャッジ日時: 2024-09-06 22:53:33
• 合計ジャッジ時間: 18,877 ms
• ジャッジサーバーID （参考情報）: judge4 / judge3

ファイルパターン	結果
sample	AC * 3
other	AC * 30

ソースコード

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import std;
void main () {
    int N = readln.chomp.to!int;
    auto x = new int[](N);
    auto y = new int[](N);
    foreach (i; 0..N) readln.read(x[i], y[i]);
    // PとQをどう求めるかが最も難しい。
    // Pについて考える。
    // 面倒なので、片方の大小関係は異なるとして考えたい。
    // xj < xiを仮定する。この時、yj < yiが必要十分条件。xの昇順にデータを追加することを考えると、rectangle 
        sumの特殊ケースに帰着する。これは平面走査で解ける。
    // Qも不等号の向きを反転させればよい。
    long P = () {
        long res = 0;
        auto index = iota(N).array.sort!((i, j) => x[i] < x[j]).array;
        auto rsq = new DynamicSegmentTree!(int, (int a, int b) => a + b, () => 0)(10L ^^ 10);
        const int pad = 10 ^^ 9;
        int L = 0, R = 0;
        while (L < N) {
            while (R < N) {
                if (x[index[L]] != x[index[R]]) break;
                R++;
            }
            foreach (U; L..R) {
                res += rsq.prod(0, y[index[U]] + pad);
            }
            // 追加
            foreach (U; L..R) {
                rsq.set(y[index[U]] + pad, rsq.get(y[index[U]] + pad) + 1);
            }
            L = R;
        }
        return res;
    }();
    long Q = () {
        long res = 0;
        auto index = iota(N).array.sort!((i, j) => x[i] < x[j]).array;
        auto rsq = new DynamicSegmentTree!(int, (int a, int b) => a + b, () => 0)(10L ^^ 10);
        const int pad = 10 ^^ 9;
        int L = 0, R = 0;
        while (L < N) {
            while (R < N) {
                if (x[index[L]] != x[index[R]]) break;
                R++;
            }
            foreach (U; L..R) {
                res += rsq.prod(y[index[U]] + pad + 1, 10L ^^ 10);
            }
            // 追加
            foreach (U; L..R) {
                rsq.set(y[index[U]] + pad, rsq.get(y[index[U]] + pad) + 1);
            }
            L = R;
        }
        return res;
    }();
    long R = () {
        long res = 0;
        auto index = iota(N).array.sort!((i, j) => x[i] < x[j]).array;
        int L = 0, R = 0;
        while (L < N) {
            while (R < N) {
                if (x[index[L]] != x[index[R]]) break;
                R++;
            }
            res += 1L * (R - L) * (N - (R - L));
            L = R;
        }
        return res / 2;
    }();
    long S = () {
        long res = 0;
        auto index = iota(N).array.sort!((i, j) => y[i] < y[j]).array;
        int L = 0, R = 0;
        while (L < N) {
            while (R < N) {
                if (y[index[L]] != y[index[R]]) break;
                R++;
            }
            res += 1L * (R - L) * (N - (R - L));
            L = R;
        }
        return res / 2;
    }();
    double ans = (P - Q) / sqrt(R.to!double) / sqrt(S.to!double);
    writefln("%.10f", ans);
}
void read (T...) (string S, ref T args) {
    import std.conv : to;
    import std.array : split;
    auto buf = S.split;
    foreach (i, ref arg; args) {
        arg = buf[i].to!(typeof(arg));
    }
}
import std.traits : ReturnType, isCallable, Parameters;
import std.meta : AliasSeq;
class DynamicSegmentTree (T, alias op, alias e) {
    // TODO: assertのメッセージを表示
    static assert(isCallable!(op));
    static assert(isCallable!(e));
    static assert(is (ReturnType!(op) == T));
    static assert(is (ReturnType!(e) == T));
    static assert(is (Parameters!(op) == AliasSeq!(T, T)));
    static assert(is (Parameters!(e) == AliasSeq!()));
    // 内部が1-indexedで動的な完全二分セグメント木
    import std.format : format;
    public:
        this (long N_)
        in { assert(1 <= N_, format("Dynamic SegmentTree: N = %s does not satisfy constraints. N must be in range of [1, %s]", 4 * 10L^^18)); }
        do {
            length = N_;
            // N_以上の2冪に設定
            N = 1;
            while (N < N_) N *= 2;
        }
        void set (long idx, T val)
        in { assert(0 <= idx && idx < length, format("Dynamic SegmentTree: idx = %s does not satisfy constraints. idx must be in range of [0, %s)", 
            idx, length)); }
        do {
            idx++;
            internal_set(root, idx, val, 1, N + 1);
        }
        T get (long idx)
        in { assert(0 <= idx && idx < length, format("Dynamic SegmentTree: idx = %s does not satisfy constraints. idx must be in range of [0, %s)", 
            idx, length)); }
        do {
            idx++;
            return internal_get(root, idx, 1, N + 1);
        }
        T prod (long l, long r)
        in {
            assert(0 <= l && l < length, format("Dynamic SegmentTree: l = %s does not satisfy constraints. l must be in range of [0, %s)", l, length
                ));
            assert(0 <= r && r <= length, format("Dynamic SegmentTree: r = %s does not satisfy constraints. r must be in range of [0, %s]", r, length
                ));
            assert(l <= r, format("Dynamic SegmentTree: l = %s, r = %s does not satisfy constraints. l <= r must be satisfied.", l, r));
        }
        do {
            l++, r++;
            if (l == r) return e();
            return internal_prod(root, l, r, 1, N + 1);
        }
        T all_prod () {
            return internal_prod(root, 1, N + 1, 1, N + 1);
        }
    private:
        struct node {
            long index;
            T value, product;
            node *left = null, right = null;
        }
        void node_update (node *n) {
            n.product = op(
                    op((n.left == null ? e() : n.left.product), n.value),
                    (n.right == null ? e() : n.right.product)
                    );
        }
        node *root = null;
        long N = 0;
        long length = 0;
        // [l, r) : 今見ている部分木が管理する範囲
        node *internal_set (ref node *cur, long idx, T val, long l, long r) {
            if (cur == null) {
                return cur = new node(idx, val, val, null, null);
            }
            if (cur.index == idx) {
                cur.value = val;
                node_update(cur);
                return cur;
            }
            // 既に部分木管理ノードが存在するときの処理
            import std.algorithm : swap;
            long mid = (l + r) / 2;
            if (idx < mid) {
                // 今いる人を押しのける
                if (cur.index < idx) { swap(cur.value, val); swap(cur.index, idx); }
                cur.left = internal_set(cur.left, idx, val, l, mid);
            }
            else {
                if (idx < cur.index) { swap(cur.value, val); swap(cur.index, idx); }
                cur.right = internal_set(cur.right, idx, val, mid, r);
            }
            node_update(cur);
            return cur;
        }
        T internal_get (const node *cur, long idx, long l, long r) {
            if (cur == null) return e();
            if (cur.index == idx) return cur.value;
            long mid = (l + r) / 2;
            if (idx < mid) return internal_get(cur.left, idx, l, mid);
            return internal_get(cur.right, idx, mid, r);
        }
        // [a, b) = 要求区間
        T internal_prod (const node *cur, long a, long b, long l, long r) {
            if (cur == null || b <= l || r <= a) return e();
            if (a <= l && r <= b) return cur.product;
            long mid = (l + r) / 2;
            T res = internal_prod(cur.left, a, b, l, mid);
            if (a <= cur.index && cur.index < b) res = op(res, cur.value);
            res = op(res, internal_prod(cur.right, a, b, mid, r));
            return res;
        }
}
 
 
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