裸的单调队列-poj-2823-Sliding Window

题目链接：

http://poj.org/problem?id=2823

题目意思：

给n个数，求连续区间长度为k的最大和最小值。

解题思路：

裸的单调队列不解释，用两个队列保存。

代码：

 

#include<iostream>

#include<cmath>

#include<cstdio>

#include<cstdlib>

#include<string>

#include<cstring>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<stack>

#include<list>

#include<queue>

#include<ctime>

#define eps 1e-6

#define INF 0x3fffffff

#define PI acos(-1.0)

#define ll __int64

#define lson l,m,(rt<<1)

#define rson m+1,r,(rt<<1)|1

#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;



#define Maxn 1100000

int sa[Maxn];

int q1[Maxn],q2[Maxn];

int ans1[Maxn],ans2[Maxn];



int main()

{

   //freopen("in.txt","r",stdin);

   //freopen("out.txt","w",stdout);

   int n,k;



   while(scanf("%d%d",&n,&k)!=EOF)

   {

       for(int i=1;i<=n;i++)

            scanf("%d",&sa[i]);

       int h1=0,t1=-1;

       int h2=0,t2=-1,cnt=0;

       for(int i=1;i<=n;i++)

       {

            while(h1<=t1&&sa[i]<=sa[q1[t1]])

                t1--;

            while(h2<=t2&&sa[i]>=sa[q2[t2]])

                t2--;

            q2[++t2]=i;

            q1[++t1]=i;

            while(q1[h1]<i-k+1)

                h1++;

            while(q2[h2]<i-k+1)

                h2++;

            if(i>=k)

            {

                ans1[++cnt]=sa[q1[h1]];

                ans2[cnt]=sa[q2[h2]];

            }



       }

       printf("%d",ans1[1]);

       for(int i=2;i<=cnt;i++)

            printf(" %d",ans1[i]);

       printf("\n%d",ans2[1]);

       for(int i=2;i<=cnt;i++)

            printf(" %d",ans2[i]);

       putchar('\n');



   }

   return 0;

}