滑动窗口

  洛谷P1886 滑动窗口
这个题与P1440 求m区间内的最小值很像，只要在这个题
基础上再另维护一个递减的单调队列就好了.

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#define inf 2147483647
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(int i=a;i<=b;++i)

using namespace std;
int n,m;
int a[2000001],q[2000001];
int top,tai;
void in(int &x){
    int y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(int x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

signed main(){
    in(n);in(m);
    For(i,1,n) in(a[i]);
    For(i,1,n){
        while(q[tai]-q[top]>=m||top==0 ) top++;
        o(a[q[top]]),p(' ');
        while(top<=tai&&a[q[tai]]>=a[i]) tai--;
        q[++tai]=i;
    }
    return 0;
}

 

 

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#define inf 2147483647
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)
//by war
//2020.3.18
using namespace std;

long long n,k;
long long a[N],q[N],b[N],s[N],Q[N];
long long l,r;
long long L,R;

void in(long long &x){
    long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

signed main()
{
    in(n);in(k);
    For(i,1,n) in(a[i]);
    For(i,1,k){
        while(q[r]-q[l]>=k ) l++;
        while(l<=r&&a[q[r]]>=a[i]) r--;
        q[++r]=i;
        
        while(Q[R]-Q[L]>=k ) L++;
        while(L<=R&&a[Q[R]]<=a[i]) R--;
        Q[++R]=i;
    }
    For(i,k+1,n+1){
        while(q[r]-q[l]>=k) l++;
        s[i]=a[q[l]];
        while(l<=r&&a[q[r]]>=a[i]) r--;
        q[++r]=i;
        
        while(Q[R]-Q[L]>=k) L++;
        b[i]=a[Q[L]];
        while(L<=R&&a[Q[R]]<=a[i]) R--;
        Q[++R]=i;
    }
    For(i,k+1,n+1) o(s[i]),p(' ');p('\n');
       For(i,k+1,n+1) o(b[i]),p(' ');p('\n');
    return 0;
}