POJ_2823_Sliding Window(RMQ / 单调队列)

Sliding Window

Time Limit: 12000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

SubmitStatus

Description

An array of size n ≤ 10 ⁶ 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

题意：有n个数，每次从左到右选取k个数，第一行选取每个区间中的最小值输出，第二行每次选取区间中的最大值输出。

分析：RMQ或单调队列求解。

RMQ：题目数据量比较大，如果用一般的二维数组显然是要MLE的，由于此题是固定的区间长度，所以只要用一维数组即可，第二维可以省略。

单调队列：数组模拟队列，用deque会超时。

题目链接：http://poj.org/problem?id=2823

代码清单：

(1) RMQ

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cstdio>
#include<string>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 1e6 + 5;

int n,k;
int a[maxn];
int minq[maxn];
int maxq[maxn];

void input(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
}

void RMQ_ST(){
    int L=(int)((log(k*1.0))/(log(2.0)));
    for(int i=1;i<=n;i++) minq[i]=maxq[i]=a[i];
    for(int j=1;j<=L;j++){
        for(int i=1;i+(1<<j)-1<=n;i++){
            minq[i]=min(minq[i],minq[i+(1<<(j-1))]);
            maxq[i]=max(maxq[i],maxq[i+(1<<(j-1))]);
        }
    }
}

void solve(){
    RMQ_ST();
    int mi=(int)((log(k*1.0))/(log(2.0)));
    for(int i=1;i+k-1<=n;i++){
        printf("%d%c",min(minq[i],minq[i+k-(1<<mi)]),i+k-1==n?'\n':' ');
    }
    for(int i=1;i+k-1<=n;i++){
        printf("%d%c",max(maxq[i],maxq[i+k-(1<<mi)]),i+k-1==n?'\n':' ');
    }
}

int main(){
    input();
    solve();
    return 0;
}

(2)单调队列

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <ctime>
#include <vector>
#include <cctype>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

#define exit() return 0

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 1e6 + 5;

struct Deque {
	int data;
	int idx;
	Deque() {}
	Deque(int _data, int _idx) : data(_data), idx(_idx) {}
};

Deque minq[maxn];
Deque maxq[maxn];
int head1, tail1;
int head2, tail2;

int n, k;
int a[maxn];
int minans[maxn];
int maxans[maxn];

void init() {
	head1 = tail1 = 0;
	head2 = tail2 = 0;
}

void input() {
	scanf("%d%d", &n, &k);
	for(int i = 0; i < n; i++) scanf("%d", &a[i]);
}

void work() {
	for(int i = 0; i < n; i++) {
		while(head1 < tail1 && minq[tail1 - 1].data > a[i]) tail1--;
		minq[tail1++] = Deque(a[i], i);
		while(head1 < tail1 && minq[head1].idx <= i - k) head1++;
		if(i >= k - 1) minans[i] = minq[head1].data;

		while(head2 < tail2 && maxq[tail2 - 1].data < a[i]) tail2--;
		maxq[tail2++] = Deque(a[i], i);
		while(head2 < tail2 && maxq[head2].idx <= i - k) head2++;
		if(i >= k - 1) maxans[i] = maxq[head2].data;
	}
}

void print() {
	for(int i = k - 1; i < n; i++) {
		if(i >= k) printf(" ");
		printf("%d", minans[i]);
	}
	printf("\n");

	for(int i = k - 1; i < n; i++) {
		if(i >= k) printf(" ");
		printf("%d", maxans[i]);
	}
	printf("\n");
}

void solve() {
	work();
	print();
}

int main() {
	init();
	input();
	solve();
	exit();
}