给一个长度为 nn 的序列 {ai}\{a_i\},定义一个区间 [l,r][l,r]的价值为这个区间中数的总和。求区间长度在 [L,R][L,R]之间的所有区间中,价值最大 kk 的个区间的价值总和。

n500,000,k500,000,1000ai1000,1LRnn\le 500,000, k\le 500,000, -1000\le a_i\le 1000, 1\le L\le R\le n

Luogu P2048

BZOJ 2006

Solution

首先预处理一下前缀和,枚举区间的右端点,可以得到一个可行的左端点区间

最有可能成为前kk大的,显然是这个区间里前缀和尽可能小的

考虑将最小的那个先算进去,如果可以是前kk大,再找第二小的,如此反复...

用堆来处理这个过程,主席树处理区间第kk

只是细节有点难调...

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
inline int read ()
{
int sum = 0, fl = 1; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
return sum * fl;
}

const int Maxn = 500000 + 100;

int N, K, L, R, Root[Maxn];
LL Sum[Maxn], A[Maxn];

namespace SEG
{
int node_cnt = 0;
struct tree
{
int ls, rs, sum;
}Tree[Maxn * 40];

inline void build (int &now, int l, int r)
{
now = ++node_cnt;
if (l == r) return ;
int mid = l + r >> 1;
build (Tree[now].ls, l, mid), build (Tree[now].rs, mid + 1, r);
}

inline void insert (int &now, int pre, int l, int r, int x)
{
now = ++node_cnt;
Tree[now].ls = Tree[pre].ls, Tree[now].rs = Tree[pre].rs;
Tree[now].sum = Tree[pre].sum + 1;
if (l == r) return ;
int mid = l + r >> 1;
if (x <= mid) insert (Tree[now].ls, Tree[pre].ls, l, mid, x);
else insert (Tree[now].rs, Tree[pre].rs, mid + 1, r, x);
}

inline int query (int now, int pre, int l, int r, int k)
{
if (l == r)
{
/**/ if (Tree[now].sum - Tree[pre].sum) return A[l];
return 0;
}
int sum = Tree[Tree[now].ls].sum - Tree[Tree[pre].ls].sum, mid = l + r >> 1;
if (k <= sum) return query (Tree[now].ls, Tree[pre].ls, l, mid, k);
return query (Tree[now].rs, Tree[pre].rs, mid + 1, r, k - sum);
}

inline int query (int x, int y, int k)
{
return query (Root[y], Root[max(0, x - 1)], 1, Maxn, k);
}
}

struct node
{
LL val;
int l, r, k, y;
bool operator < (const node &x) const { return x.val > val; }
};

priority_queue <node> Q;

int main()
{
#ifdef hk_cnyali
freopen("A.in", "r", stdin);
freopen("A.out", "w", stdout);
#endif
N = read(), K = read(), L = read(), R = read();
/**/++N;
for (int i = 2; i <= N; ++i)
A[i] = Sum[i] = Sum[i - 1] + read();
sort(A + 1, A + N + 1);
int len = unique(A + 1, A + N + 1) - A - 1;
SEG :: build (Root[0], 1, Maxn);
for (int i = 1; i <= N; ++i)
{
int pos = lower_bound (A + 1, A + len + 1, Sum[i]) - A;
SEG :: insert (Root[i], Root[i - 1], 1, Maxn, pos);
}
for (int i = L + 1; i <= N; ++i)
{
/**/ int l = max(0, i - R), r = max(0, i - L);
Q.push((node){Sum[i] - SEG :: query (l, r, 1), l, r, 1, i});
}
LL ans = 0;
while (K-- && !Q.empty())
{
node x = Q.top(); Q.pop();
ans += x.val;
int len_now = x.r - x.l + 1;
/**/ if (!x.l) len_now--;
if (x.k == len_now) continue;
Q.push((node){Sum[x.y] - SEG :: query (x.l, x.r, x.k + 1), x.l, x.r, x.k + 1, x.y});
}
cout<<ans<<endl;
return 0;
}

Debug

  • 61L: 如果区间根本不存在值return 0

  • 91L: 要在最前面多加一个把00的位置++的线段树,因为前缀和也可能是0

  • 104L: 前缀和是减去suml1sum_{l-1}而不是sumlsum_l

  • 113L: 特判一下0的情况...