poj2823 单调队列

//poj2823
//维护两个单调队列
//单调队列队头最大或最小，值和值的序号都要单调！
#include <iostream>
#include <queue>
using namespace std;
struct data {
int x,id;
}q1[1000010],q2[1000010];
int ans1[1000010],ans2[1000010],a[1000010],n,k,i;
int head1,tail1,head2,tail2;
void insert1(int x,int id) {
if (head1==tail1+1) {
q1[head1].x=x;
q1[head1].id=id;
tail1++;
}
else {
while (q1[tail1].x>x && head1<=tail1) tail1--;
tail1++;
q1[tail1].x=x;
q1[tail1].id=id;
}
}
void insert2(int x,int id) {
if (head2==tail2+1) {
q2[head2].x=x;
q2[head2].id=id;
tail2++;
}
else {
while (q2[tail2].x<x && head2<=tail2) tail2--;
tail2++;
q2[tail2].x=x;
q2[tail2].id=id;
}
}
int main() {
cin >> n >> k;
for (i=1;i<=n;i++)
scanf("%d",&a[i]);
head1=1;tail1=0;
head2=1;tail2=0;
for (i=1;i<=k;i++) {
insert1(a[i],i);
insert2(a[i],i);
}
ans1[1]=q1[1].x;
ans2[1]=q2[1].x;
for (i=1;i<=n-k;i++) {
if (q1[head1].id==i) head1++;
if (q2[head2].id==i) head2++;
insert1(a[i+k],i+k);
insert2(a[i+k],i+k);
ans1[i+1]=q1[head1].x;
ans2[i+1]=q2[head2].x;
}
for (i=1;i<n-k+1;i++)
printf("%d ",ans1[i]);
printf("%d/n",ans1[n-k+1]);
for (i=1;i<n-k+1;i++)
printf("%d ",ans2[i]);
printf("%d/n",ans2[n-k+1]);
//system("pause");
return 0;
}