【BZOJ4653】【UOJ222】【LOJ2086】【NOI2016】区间（线段树）

Description

http://uoj.ac/problem/222

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Solution

线段树水题。

将线段按长度排序，由于花费等于最长线段减最短线段，我们可以只考虑选取一段连续区间内的线段。
我们对于每一个左端点，找到一个最靠左的右端点计算贡献。而右端点的位置是单调的，我们可以做到$O(n)$枚举，加上线段树，时间复杂度$O(n\log n)$。

加上标记永久化比pushdown快了不少诶。

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Code

/**************************************
 * Au: Hany01
 * Prob: [BZOJ4653][UOJ222][NOI2016] 区间
 * Date: Jul 31st, 2018
 * Email: [email protected]
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = j; i < i##_end_; ++ i)
#define For(i, j ,k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define x first
#define y second
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    register char c_; register int _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 5e5 + 5, maxr = maxn << 1;

struct Seg {
    int l, r, len;
    bool operator < (const Seg& rhs) const { return len < rhs.len; }
}seg[maxn];

int tr[maxr << 2], tag[maxr << 2];

#define mid ((l + r) >> 1)
#define lc  (t << 1)
#define rc  (lc | 1)

void update(int t, int l, int r, int x, int y, int dt) {
    if (x <= l && r <= y) { tag[t] += dt, tr[t] += dt; return; }
    if (x <= mid) update(lc, l, mid, x, y, dt);
    if (y >  mid) update(rc, mid + 1, r, x, y, dt);
    tr[t] = max(tr[lc], tr[rc]) + tag[t];
}

int main()
{
#ifdef hany01
    File("bzoj4653");
#endif

    static int n, m, Ans, ls[maxr], tot, R;

    n = read(), m = read();
    For(i, 1, n) seg[i].l = ls[++ tot] = read(), seg[i].r = ls[++ tot] = read(), seg[i].len = seg[i].r - seg[i].l;
    stable_sort(ls + 1, ls + 1 + tot), tot = unique(ls + 1, ls + 1 + tot) - ls - 1;
    For(i, 1, n)
        seg[i].l = lower_bound(ls + 1, ls + 1 + tot, seg[i].l) - ls,
        seg[i].r = lower_bound(ls + 1, ls + 1 + tot, seg[i].r) - ls;
    stable_sort(seg + 1, seg + 1 + n);

    Ans = INF1;
    For(L, 1, n) {
        while (tr[1] < m && R < n) ++ R, update(1, 1, tot, seg[R].l, seg[R].r, 1);
        if (tr[1] < m) break;
        chkmin(Ans, seg[R].len - seg[L].len);
        update(1, 1, tot, seg[L].l, seg[L].r, -1);
    }

    printf("%d\n", Ans == INF1 ? -1 : Ans);

    return 0;
}