「题解」洛谷 P2234 [HNOI2002]营业额统计

题目

P2234 [HNOI2002]营业额统计

简化题意

给你一个长为 (n) 的序列，让你求 (sumlimits_{i = 1} ^ {n} min (abs(a_j - a_i)),j in [1, i - 1))

思路

Splay。

对于一颗建好的 Splay 每次插入 (a_i)，最小的贡献只有可能出现在前驱和后继中。

Code

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#define MAXN 33333

int min(int a, int b) { return a < b ? a : b; }

inline void read(int &T) {
    int x = 0;
    bool f = 0;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = !f;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    T = f ? -x : x;
}

int n, ans, root, fa[MAXN], son[MAXN][2];
int num, size[MAXN], val[MAXN], cnt[MAXN];

namespace Splay {
    int which(int x) { return son[fa[x]][1] == x ? 1 : 0; }
    void update(int x) { size[x] = cnt[x] + size[son[x][0]] + size[son[x][1]]; }
    void clear(int x) { fa[x] = son[x][0] = son[x][1] = cnt[x] = size[x] = val[x] = 0; }
    void rotate(int x) {
        int father = fa[x], grandpa = fa[father];
        int flag1 = which(x), flag2 = which(father);
        fa[x] = grandpa, fa[father] = x;
        if (grandpa) son[grandpa][flag2] = x;
        fa[son[x][flag1 ^1]] = father;
        son[father][flag1] = son[x][flag1 ^1];
        son[x][flag1 ^ 1] = father;
        update(x), update(father);
    }
    void splay(int x) {
        for (int f = fa[x]; f = fa[x], f; rotate(x)) {
            if (fa[f]) rotate(which(x) == which(f) ? f : x);
        }
        root = x;
    }
    int pre() {
        int now = son[root][0];
        while (son[now][1]) now = son[now][1];
        return now;
    }
    int nxt() {
        int now = son[root][1];
        while (son[now][0]) now = son[now][0];
        return now;
    }
    void ins(int x) {
        if (!root) {
            val[++num] = x;
            ++cnt[num];
            root = num;
            update(num);
            return;
        }
        int now = root, f = 0;
        while (1) {
            if (val[now] == x) {
                ++cnt[now];
                update(now), update(f);
                splay(now);
                return;
            }
            f = now, now = son[now][val[now] < x];
            if (!now) {
                val[++num] = x;
                ++cnt[num];
                fa[num] = f;
                son[f][val[f] < x] = num;
                update(num), update(f);
                splay(num);
                return;
            }
        }
    }
}

int main() {
    read(n);
    for (int i = 1, x; i <= n; ++i) {
        read(x);
        Splay::ins(x);
        if (i == 1) ans += x;
        else if (cnt[root] > 1) continue;
        else {
            int minn = 2147483647, maxx = 2147483647;
            if (son[root][0]) minn = val[Splay::pre()];
            if (son[root][1]) maxx = val[Splay::nxt()];
            ans += min(abs(minn - x), abs(maxx - x));
        }
    }
    std::cout << ans << '
';
    return 0;
}