poj2823 Sliding Window

来源：互联网发布：软件系统架构图编辑：程序博客网时间：2024/05/22 17:10

单调队列维护（以求最大值为例：每次现在队尾维护，删除掉比<=x的元素，插入x，因为删除的元素比x小而且比x出的早，没有用了！然后再在队首维护，因为是按下表从小到大排列的，只需把队首该出队的删除了，剩下的队首就是答案）

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#define N 1000050using namespace std;struct xxx{    int pos,v;}que1[N],que2[N];int head1,tail1,head2,tail2,maxn[N],minn[N];int main(){    memset(que1,0,sizeof(que1));    memset(que2,0,sizeof(que2));    memset(maxn,0,sizeof(maxn));    memset(minn,0,sizeof(minn));    head1=head2=0;    tail1=tail2=-1;    int n,k,x,temp;    scanf("%d%d",&n,&k);    for(int i=1;i<=n;i++)    {        scanf("%d",&x);        while(head1<=tail1&&que1[tail1].v<=x)tail1--;        que1[++tail1].v=x;que1[tail1].pos=i;        temp=i-k;        if(temp<0)temp=0;        while(que1[head1].pos<=temp)head1++;        maxn[i]=que1[head1].v;        while(head2<=tail2&&que2[tail2].v>=x)tail2--;        que2[++tail2].v=x;que2[tail2].pos=i;        temp=i-k;        if(temp<0)temp=0;        while(que2[head2].pos<=temp)head2++;        minn[i]=que2[head2].v;    }    for(int i=k;i<n;i++)         printf("%d ",minn[i]);    printf("%d\n",minn[n]);    for(int i=k;i<n;i++)         printf("%d ",maxn[i]);    printf("%d\n",maxn[n]);    return 0;}

PS：C++ AC了，G++ TLE了。。。- -#为神马？