CF371E Subway Innovation 题解

合集 - codeforces题解集1(37)

1.CF1916E Happy Life in University 题解2023-12-312.CF763E Timofey and our friends animals题解2024-01-033.CF1270G Subset with Zero Sum2024-01-054.CF1045G AI robots题解2024-01-055.CF940F Machine Learning题解2024-01-066.CF678F Lena and Queries题解2024-01-127.CF1921F Sum of Progression 题解2024-01-168.CF526F Pudding Monsters 题解2024-01-219.CF455D Serega and Fun 题解2024-01-2210.CF351D Jeff and Removing Periods 题解2024-01-2211.CF452F Permutation 与 P2757 [国家集训队] 等差子序列 题解2024-01-2312.CF911G Mass Change Queries 题解2024-01-2413.CF145E Lucky Queries 题解2024-01-2514.CF1515F Phoenix and Earthquake 题解2024-01-2615.CF765F Souvenirs 题解2024-01-2716.CF1764H Doremy's Paint 2 题解2024-01-2917.CF1000F One Occurrence题解2024-01-3018.CF813E Army Creation 题解2024-01-3119.CF1706E Qpwoeirut and Vertices 题解2024-02-0120.CF620E New Year Tree 题解2024-02-0221.CF1454F Array Partition 题解2024-02-0322.CF1771F Hossam and Range Minimum Query 题解2024-02-0823.CF1931F Chat Screenshots 另一种题解2024-02-1424.CF896C Willem, Chtholly and Seniorious 题解2024-02-1525.CF55D Beautiful numbers 题解2024-02-1626.CF916E Jamie and Tree 题解2024-02-2327.CF696B Puzzles 题解2024-02-2828.CF383C Propagating tree 题解2024-02-2929.CF1436E Complicated Computations 题解2024-03-0830.CF817F MEX Queries 题解2024-03-2031.CF1638E Colorful Operations 题解2024-03-2132.CF1618G Trader Problem 题解2024-03-2233.CF794F Leha and security system 题解2024-03-23

34.CF371E Subway Innovation 题解2024-04-04

35.CF1200E Compress Words 题解2024-04-0636.CF1884D Counting Rhyme 题解2024-05-1737.CF1982F Sorting Problem Again 题解2024-06-26

收起

题目链接：CF 或者 洛谷

对于绝对值的几何意义来说，这题是在直线上的两点间的距离，为了总的距离和最下，首先最好让它们两两之间最好都紧挨着。

由于询问的是 $(i,j)$ 不重不漏的对有关，即 $(i<j)+(i>j)+(i=j)=all(i,j)$，又因为，$(i,j)$ 的贡献和 $(j,i)$ 相同且重复，所以我们只需要加一个偏序关系就行了。

考虑一下题目让求的东西：

$$min\sum _{i=1}^k\sum _{j=1}^{i-1}|x_i-x_j|$$

先考虑去绝对值。其实只需要让 $x$ 排序就行，因为 $j<i$，所以 $x_i\ge x_j$，绝对值就就能去掉了。

$$min\sum _{i=1}^k\sum _{j=1}^{i-1}{x}_i-x_j$$

这玩意可以看做从 $n$ 个点排序后中取得一个 $k$ 个点的子序列，算上述式子最佳答案的 $k$ 个点。

每选择一个点，假设它是第 $t+1$ 个点，那么我们发现，会增加这第 $t+1$ 个点与前面 $t$ 个点距离之和的贡献，越往左走，这个贡献越小，最左不能超过第 $t$ 个点，所以我们有贪心策略，让这 $k$ 个点尽量挨着，从子序列的选择，变为了子数组的选择。

接下来其实就类似滑动窗口，枚举起点，算出贡献，比较即可。

先考虑多一个数的贡献：

$$\sum _{j=l}^{r-1}{x}_r-x_j=(r-l)\times x_r-\sum _{j=l}^{r-1}{x}_j$$

$$\mathrm{前}\mathrm{缀}\mathrm{和}\mathrm{优}\mathrm{化}:\sum _{j=l}^{r-1}{x}_j=pre[r-1]-pre[l-1]$$

少一个数的贡献：

$$\mathrm{窗}\mathrm{口}\mathrm{满}\mathrm{时}\mathrm{移}\mathrm{动}:\sum _{j=l+1}^r{x}_j-x_l=\sum _{j=l+1}^r{x}_j-(k-1)\times x_l$$

$$\mathrm{前}\mathrm{缀}\mathrm{和}\mathrm{优}\mathrm{化}:\sum _{j=l+1}^r{x}_j=pre[r]-pre[l]$$

所以滑窗转移就是 $O(1)$，就做完了。

参照代码

#include <bits/stdc++.h>
 
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
 
// #define isPbdsFile
 
#ifdef isPbdsFile
 
#include <bits/extc++.h>
 
#else
 
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
 
#endif
 
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
 
template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}
 
template <typename T>
T lowBit(T x)
{
    return x & -x;
}
 
template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}
 
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}
 
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c) if (b & 1) (ans *= a) %= c;
    return modt(ans, c);
}
 
template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-') sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}
 
template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}
 
template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char)) return;
    if (x < 0) x = -x, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 ^ 48);
}
 
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}
 
 
template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;
 
    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow) return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }
 
    T3()
    {
        one = tow = three = 0;
    }
 
    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};
 
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y) x = y;
}
 
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y) x = y;
}
 
constexpr int N = 3e5 + 10;
pll a[N];
int n, k;
ll pre[N];
ll curr, mi = LLONG_MAX;
int ansIdx;
 
inline void solve()
{
    cin >> n;
    forn(i, 1, n) cin >> a[i].first, a[i].second = i;
    sortArr(a, n);
    forn(i, 1, n) pre[i] = pre[i - 1] + a[i].first;
    cin >> k;
    ll l = 1;
    forn(r, 1, n)
    {
        curr += (r - l) * a[r].first - (pre[r - 1] - pre[l - 1]);
        if (r >= k)
        {
            if (curr < mi) mi = curr, ansIdx = l;
            curr -= pre[r] - pre[l] - (k - 1) * a[l++].first;
        }
    }
    while (k--) cout << a[ansIdx++].second << ' ';
}
 
signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test) solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

$$\mathrm{时}\mathrm{间}\mathrm{复}\mathrm{杂}\mathrm{度}\mathrm{为}:O(n\log n)$$