P2698 [USACO12MAR]花盆Flowerpot 单调队列

Posted 2021-02-09 ckxkexing

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>

/*
        
⊂_ヽ
　 ＼＼ Λ＿Λ  来了老弟
　　 ＼(‘?‘)
　　　 >　⌒ヽ
　　　/ 　 へ＼
　　 /　　/　＼＼
　　 ?　ノ　　 ヽ_つ
　　/　/
　 /　/|
　(　(ヽ
　|　|、＼
　| 丿 ＼ ⌒)
　| |　　) /
‘ノ )　　L?

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#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 ‘
‘

#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(‘
‘);
    }
    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
", ans);
            return 0;
}

View Code