POJ 2823 Sliding Window 再探单调队列

重新刷这个经典题，感觉跟以前不一样了，变得更加容易理解了，不讲解了，看代码。注意：要用C++提交，用G++会超时。。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

#define N 1000007



int a[N],mp[N],head,tail,n,k;



inline void pushup(int i)

{

    while(tail > head && a[i] < a[mp[tail-1]])

        tail--;

    mp[tail++] = i;  //mq 

}



inline void pushdown(int i)

{

    while(tail > head && a[i] > a[mp[tail-1]])

        tail--;

    mp[tail++] = i;

}



void solve(int cmd)

{

    head = tail = 0;

    int i;

    void (*push)(int) = cmd?pushdown:pushup;

    for(i=1;i<=k&&i<=n;i++)

        push(i);

    printf("%d",a[mp[head]]);

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

    {

        while(head != tail && mp[head] < i-k+1)

            head++;

        push(i);

        printf(" %d",a[mp[head]]);

    }

    printf("\n");

}



int main()

{

    int i;

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

    {

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

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

        solve(0);

        solve(1);

    }

    return 0;

}

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