USACO 2013 November Contest, Silver Problem 2. Crowded Cows 单调队列

Problem 2: Crowded Cows [Brian Dean, 2013]

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.

PROBLEM NAME: crowded

INPUT FORMAT:

* 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.

SAMPLE INPUT (file crowded.in):

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.

OUTPUT FORMAT:

* Line 1: The number of crowded cows.

SAMPLE OUTPUT (file crowded.out):

2 

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

分析

首先不难想到先按照奶牛的位置进行排序。

这道题的关键在于维护一个单调队列。当一个新的奶牛需要入队时，先和队尾进行比较。如果队尾奶牛高度小于需要入队的奶牛高度，那么删除队尾。直到满足队尾奶牛高度大于需要入队的奶牛高度。（因为如果后面的奶牛感到拥挤，有了这个入队的奶牛，这些被删除的奶牛已经不再有必要性了）

插入这个奶牛时，再与队头的奶牛比较删除队头奶牛，直到插入奶牛与队头奶牛之间的距离小于等于D。（因为距离大于D的奶牛已经不会再对之后的奶牛造成影响了）

最后比较插入奶牛的身高与队头奶牛身高（此时队头奶牛身高最高且距离小于等于D）。如果满足大于两倍身高那么说明在这个奶牛的左面已经感到拥挤了。

然后把上述过程反过来从右向左维护单调队列。

最后从1~n的奶牛中寻找，如果奶牛在左右两侧都有满足要求会使它感到拥挤的奶牛，那么这是一个合法情况，将计数器加1。

我自己写程序犯了一些小错误：把某两个cows写成了Q。所以同类型的变量写的时候还是过一下脑子自己在写什么蛤。

程序

#include <bits/stdc++.h>
using namespace std;
const int MAX = 50000 + 1;
int n, d;
struct cow
{
    int x,h;
}cows[MAX];
inline bool cmp(cow a, cow b)
{
    return a.x < b.x;
}
int main()
{
    cin >> n >> d;
    for (int i = 1; i <= n; i++)
        cin >> cows[i].x >> cows[i].h;
    sort(cows+1, cows+(n+1), cmp);
    cow Q[MAX];
    int Head = 1, Tail = 0;
    bool vis[MAX], vis2[MAX];
    for(int i = 1; i <= n; i++)
    {
        while(Head <= Tail && Q[Tail].h < cows[i].h)
            Tail--;
        Q[++Tail] = cows[i];
        while(Head <= Tail && Q[Head].x < cows[i].x-d)
            Head++;
        if(Q[Head].h >= cows[i].h*2)
            vis[i]=true;
    }
    memset(Q, 0, sizeof(Q));
    Head=1;
    Tail=0;
    for(int i = n; i >= 1; i--)
    {
        while(Head <= Tail && Q[Tail].h < cows[i].h)
            Tail--;
        Q[++Tail] = cows[i];
        while(Head <= Tail && Q[Head].x > cows[i].x+d)
            Head++;
        if(Q[Head].h >= cows[i].h*2)
            vis2[i]=true;
    }
    int ans=0;
    for(int i = 1; i <= n; i++)
    {
        if(vis[i] && vis2[i])
            ans++;
        //cout << vis[i] << vis2[i] << endl;
    }
    printf("%d",ans);
    return 0;
}