从左到右扫一遍, 维护一个单调不递减队列. 然后再从右往左重复一遍然后就可以统计答案了。

----------------------------------------------------------------------------

#include<bits/stdc++.h>
 
#define rep(i, n) for(int i = 0; i < n; ++i)
#define clr(x, c) memset(x, c, sizeof(x))
#define foreach(i, x) for(__typeof(x.begin()) i = x.begin(); i != x.end(); i++)
 
using namespace std;
 
const int maxn = 50009;
 
struct R {
int p, h;
inline void Read() {
scanf("%d%d", &p, &h);
}
bool operator < (const R &t) const {
return p < t.p;
}
} A[maxn];
 
bool l[maxn], r[maxn];
int n, d;
deque<int> Q;
 
int main() {
freopen("test.in", "r", stdin);
cin >> n >> d;
rep(i, n) A[i].Read();
sort(A, A + n);
clr(l, 0), clr(r, 0);
while(!Q.empty()) Q.pop_back();
rep(i, n) {
R* h = A + i;
while(!Q.empty() && A[Q.front()].p + d < h->p) Q.pop_front();
if(!Q.empty() && A[Q.front()].h >= 2 * h->h) l[i] = true;
while(!Q.empty() && A[Q.back()].h < h->h) Q.pop_back();
Q.push_back(i);
}
while(!Q.empty()) Q.pop_back();
for(int i = n - 1; ~i; i--) {
R* h = A + i;
while(!Q.empty() && A[Q.front()].p - d > h->p) Q.pop_front();
if(!Q.empty() && A[Q.front()].h >= 2 * h->h) r[i] = true;
while(!Q.empty() && A[Q.back()].h < h->h) Q.pop_back();
Q.push_back(i);
}
int ans = 0;
rep(i, n) if(l[i] && r[i]) ans++;
cout << ans << "\n";
return 0;
}

----------------------------------------------------------------------------

3314: [Usaco2013 Nov]Crowded Cows

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 120  Solved: 85
[Submit][Status][Discuss]

Description

Farmer John's N cows (1 <= N <= 50,000) are grazing along a one-dimensional fence. Cow i is standing at location x(i) and has height h(i) (1 <= x(i),h(i) <= 1,000,000,000). A cow feels "crowded" if there is another cow at least twice her height within distance D on her left, and also another cow at least twice her height within distance D on her right (1 <= D <= 1,000,000,000). Since crowded cows produce less milk, Farmer John would like to count the number of such cows. Please help him.

N头牛在一个坐标轴上,每头牛有个高度。现给出一个距离值D。

如果某头牛在它的左边,在距离D的范围内,如果找到某个牛的高度至少是它的两倍,且在右边也能找到这样的牛的话。则此牛会感觉到不舒服。

问有多少头会感到不舒服。

Input

* Line 1: Two integers, N and D.

* Lines 2..1+N: Line i+1 contains the integers x(i) and h(i). The locations of all N cows are distinct.

Output

* Line 1: The number of crowded cows.

Sample Input

6 4
10 3
6 2
5 3
9 7
3 6
11 2

INPUT DETAILS: There are 6 cows, with a distance threshold of 4 for feeling crowded. Cow #1 lives at position x=10 and has height h=3, and so on.

Sample Output

2
OUTPUT DETAILS: The cows at positions x=5 and x=6 are both crowded.

HINT

Source

BZOJ 3314: [Usaco2013 Nov]Crowded Cows( 单调队列 )的更多相关文章

  1. BZOJ 3314 [Usaco2013 Nov]Crowded Cows:单调队列

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=3314 题意: N头牛在一个坐标轴上,每头牛有个高度.现给出一个距离值D. 如果某头牛在它的 ...

  2. 【BZOJ3314】 [Usaco2013 Nov]Crowded Cows 单调队列

    第一次写单调队列太垃圾... 左右各扫一遍即可. #include <iostream> #include <cstdio> #include <cstring> ...

  3. 3314: [Usaco2013 Nov]Crowded Cows

    3314: [Usaco2013 Nov]Crowded Cows Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 111  Solved: 79[Sub ...

  4. 【BZOJ】3314: [Usaco2013 Nov]Crowded Cows(单调队列)

    http://www.lydsy.com/JudgeOnline/problem.php?id=3314 一眼就是维护一个距离为d的单调递减队列... 第一次写.....看了下别人的代码... 这一题 ...

  5. BZOJ3314: [Usaco2013 Nov]Crowded Cows

    3314: [Usaco2013 Nov]Crowded Cows Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 86  Solved: 61[Subm ...

  6. BZOJ : [Usaco2013 Nov]Crowded 单调队列

    正反两遍个来一次单调队列 DP 即可. Code: #include<cstdio> #include<deque> #include<algorithm> usi ...

  7. 【bzoj 1414】对称的正方形 单调队列+manacher

    Description Orez很喜欢搜集一些神秘的数据,并经常把它们排成一个矩阵进行研究.最近,Orez又得到了一些数据,并已经把它们排成了一个n行m列的矩阵.通过观察,Orez发现这些数据蕴涵了一 ...

  8. bzoj 1047 : [HAOI2007]理想的正方形 单调队列dp

    题目链接 1047: [HAOI2007]理想的正方形 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 2369  Solved: 1266[Submi ...

  9. BZOJ 2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛( dp )

    树形dp..水 ------------------------------------------------------------------------ #include<cstdio& ...

随机推荐

  1. bootstrap基础知识点YI

    <!DOCTYPE html> <html lang="en"> ... </html> bootstrap页面都应该包含html5声明. 框架 ...

  2. Python之路:Python 函数

    一.函数式编程:将某功能代码封装到函数中,日后便无需重复编写,仅调用函数即可 面向对象:对函数进行分类和封装 二. 函数的定义和使用 def 函数名(参数): ... 函数体 ... 函数的定义主要有 ...

  3. 【转】C++常见错误大全

    原文转自:http://hi.baidu.com/qiou2719/item/b9eed949130ff50ec0161331 C++常见错误大全 0. XXXX "is not a cla ...

  4. Log Collect

    http://ossectools.blogspot.com/2011/03/comprehensive-log-collection.html https://www.hacking-lab.com ...

  5. poj 2752 Seek the Name, Seek the Fame(KMP需转换下思想)

    Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10204   Ac ...

  6. <jsp:include page="">和<%@include page=""%> 标签学习

    <jsp:include page=""><jsp:param value=""name=""/><DEL&g ...

  7. Objective中的协议(Protocol)

    Objective中的协议(Protocol) 作用: 专门用来声明一大堆方法. (不能声明属性,也不能实现方法,只能用来写方法的声明). 只要某个类遵守了这个协议.就相当于拥有这个协议中的所有的方法 ...

  8. Button的设置及各种属性

    (1)UIButton类继承自UIControl,而UIControl继承自UIView,因为UIView就是个矩形区域,所以UIButton实例化的对象其实都是一个矩形,虽然有各种圆角.增加联系人. ...

  9. 怎么在Ubuntu Scope中获取location地址信息

    Location信息对非常多有地址进行搜索的应用来说非常重要.比方对dianping这种应用来说.我们能够通过地址来获取当前位置的一些信息.在这篇文章中,我们来介绍怎样获取Scope架构中的位置信息. ...

  10. BZOJ 2843: 极地旅行社( LCT )

    LCT.. ------------------------------------------------------------------------ #include<cstdio> ...