P2698 [USACO12MAR]花盆Flowerpot 单调队列

https://www.luogu.org/problemnew/show/P2698

警示

用数组写双端队列的话，记得le = 1， ri = 0；
le<=ri表示队列非空

题意

求一个最小的区间长度，使得区间中的最大值和最小值的差>=D.

思路

一开始二分加线段树强行做，多了一个log。用ST表可能会优秀。做到nlogn。
但是如果用单调队列的话，除去排序，就可以做到O（n）
具体来说，对于一个L，合法的最小的右区间若为R，那么L+1的最小合法右区间一定>=R.

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
struct FastIO {
    static const int S = 4e6;
    int wpos;
    char wbuf[S];
    FastIO() :
        wpos(0) {
    }
    inline int xchar() {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len) exit(0);
        return buf[pos++];
    }
    inline int xuint() {
        int c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x;
    }
    inline int xint() {
        int s = 1, c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        if (c == '-') s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s) {
        int c = xchar();
        while (c <= 32) c = xchar();
        for (; c > 32; c = xchar()) *s++ = c;
        *s = 0;
    }
    inline void wchar(int x) {
        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos++] = x;
    }
    inline void wint(int x) {
        if (x < 0) wchar('-'), x = -x;
        char s[24];
        int n = 0;
        while (x || !n) s[n++] = '0' + x % 10, x /= 10;
        while (n--) wchar(s[n]);
        wchar('\n');
    }
    inline void wstring(const char *s) {
        while (*s) wchar(*s++);
    }
    ~FastIO() {
        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;
inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/
const int maxn = 100009;
struct node {
    int x, y;
} a[maxn];
bool cmp(node a, node b) {
    return a.x < b.x;
}
int dq1[maxn * 2], dq2[maxn * 2];
int main() {
    int n, D;
    scanf("%d%d", &n, &D);
    rep(i, 1, n) {
        scanf("%d%d", &a[i].x, &a[i].y);
    }
    sort(a + 1, a + 1 + n, cmp);
    int ans = inf;

    int l1 = 1, r1 = 0, l2 = 1, r2 = 0;
    for (int i = 1, r = 0; i <= n; i++) {
        while (l1 <= r1 && dq1[l1] < i) l1++;
        while (l2 <= r2 && dq2[l2] < i) l2++;

        while (a[dq1[l1]].y - a[dq2[l2]].y < D && r < n) {
            r++;
            while (a[dq1[r1]].y <= a[r].y && l1 <= r1) r1--;
            dq1[++r1] = r;
            while (a[dq2[r2]].y >= a[r].y && l2 <= r2) r2--;
            dq2[++r2] = r;
        }

        if (l1 <= r1 && l2 <= r2 && a[dq1[l1]].y - a[dq2[l2]].y >= D) ans = min(ans, a[r].x - a[i].x);
    }
    if (ans >= inf)
        puts("-1");
    else
        printf("%d\n", ans);
    return 0;
}