【单调队列】POJ 2823 Sliding Window

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求连续的K个元素中的最小值和最大值

#include <cstdio>#include <cstring>#include <cstdlib>#include <string>#include <iostream>#include <algorithm>#include <sstream>#include <cmath>using namespace std;#include <queue>#include <stack>#include <vector>#include <deque>#define cler(arr, val)    memset(arr, val, sizeof(arr))typedef long long  LL;const int MAXN = 1002000;const int MAXM = 6000010;const int INF = 0x3f3f3f3f;const int mod = 1000000007;int a[MAXN],q[MAXN];int n,k;void getnum1()//取最小{    int l=1,r=0;//l    cler(q,0);    q[1]=1;    for(int i=1;i<=n;i++)    {        if(i-q[l]==k) l++;        if(a[q[r]]<a[i]||r==l-1)//将第i个放进队尾            q[++r]=i;        else//更新队尾的元素 一直到比第i个还要小        {            while(r>=l&&a[i]<=a[q[r]])                q[r--]=i;            r++;        }        if(i>=k)printf("%d ",a[q[l]]);    }    puts("");}void getnum2(){    int l=1,r=0;    cler(q,0);    q[1]=1;    for(int i=1;i<=n;i++)    {        if(i-q[l]==k) l++;        if(a[q[r]]>a[i]||r==l-1)            q[++r]=i;        else        {            while(r>=l&&a[i]>=a[q[r]])                q[r--]=i;            r++;        }        if(i>=k)printf("%d ",a[q[l]]);    }    puts("");}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    // freopen("out.txt", "w", stdout);#endif    while(cin>>n>>k)    {        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        getnum1();        getnum2();    }    return 0;}

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