POJ2823 Sliding Window(單調隊列)

題目：

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 48443		Accepted: 13988
Case Time Limit: 5000MS

Description

An array of size n ≤ 10⁶ is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.

Window position	Minimum value	Maximum value
[1  3  -1] -3  5  3  6  7	-1	3
1 [3  -1  -3] 5  3  6  7	-3	3
1  3 [-1  -3  5] 3  6  7	-3	5
1  3  -1 [-3  5  3] 6  7	-3	5
1  3  -1  -3 [5  3  6] 7	3	6
1  3  -1  -3  5 [3  6  7]	3	7

Your task is to determine the maximum and minimum values in the sliding window at each position.

Input

The input consists of two lines. The first line contains two integers n and k which are the lengths of the array and the sliding window. There are n integers in the second line.

Output

There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.

Sample Input

8 3
1 3 -1 -3 5 3 6 7

Sample Output

-1 -3 -3 -3 3 3
3 3 5 5 6 7

Source

POJ Monthly--2006.04.28, Ikki

題意：給一個長度爲n的序列，求出其中每個長度爲k的區間的最大值和最小值。

思路：我們可以用兩個單調隊列來分別維護最大值和最小值，先使前k個元素入隊，找到其中的最大值和最小值（即隊首元素），然後把後面的元素一次入隊，每一個元素入隊就代表了一個新的長度爲k的區間，然後我們找出跟當前位置的距離在k以內的隊首保存下來即可。
代碼：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判斷正負
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的數
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/


struct node
{
    int num;
    int po;
};
int a[1000000+5];
node q1[1000000+5];
node q2[1000000+5];
int ans1[1000000+5];
int ans2[1000000+5];
int main()
{int n,k;
RII(n,k);
for(int i=0;i<n;i++)
    RI(a[i]);
int front1=0,rear1=0,front2=0,rear2=0;
for(int i=0;i<k&&i<n;i++)
{while(front1<rear1&&q1[rear1-1].num>=a[i])
rear1--;
q1[rear1].num=a[i];
q1[rear1++].po=i;
while(front2<rear2&&q2[rear2-1].num<=a[i])
    rear2--;
q2[rear2].num=a[i];
q2[rear2++].po=i;
}
int p1=0,p2=0;
ans1[p1++]=q1[front1].num;
ans2[p2++]=q2[front2].num;
for(int i=k;i<n;i++)
{
    while(front1<rear1&&q1[rear1-1].num>=a[i])
rear1--;
q1[rear1].num=a[i];
q1[rear1++].po=i;
while(front2<rear2&&q2[rear2-1].num<=a[i])
    rear2--;
q2[rear2].num=a[i];
q2[rear2++].po=i;
while(i-q1[front1].po>=k)
    front1++;
ans1[p1++]=q1[front1].num;
while(i-q2[front2].po>=k)
    front2++;
ans2[p2++]=q2[front2].num;
}
for(int i=0;i<p1-1;i++)
    printf("%d ",ans1[i]);
printf("%d\n",ans1[p1-1]);
for(int i=0;i<p2-1;i++)
    printf("%d ",ans2[i]);
printf("%d\n",ans2[p2-1]);

        return 0;
}