解题报告：Fake Maxpooling（单调队列求矩阵的和）

我们不妨先把这个问题中二维的矩阵简化成一维的数列。那么现在的问题就变成了一个求连续的滑动窗口最值问题：给出一个长度为n的数列和一个长度为k(k$O((n-k+1)\ast k)$

1

#include
#include
#include
#include
#include
#include
using namespace std;
 
typedef long long ll;
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f;
const int N = 5e3+7;
int n,m,k;
int a[N][N],b[N][N];
int f[N];
int que[N];
void getmaxa(int x,int len){
    memset(que,0,sizeof que);
    int i,head = 1,tail = 1;
    for(i = 1;i < k;++i){
        while(tail >= head && f[i] > f[que[tail]])
            tail--;
        tail++;
        que[tail] = i;
    }
    for(i = k;i <= len;++i){
        while(tail >= head && i - que[head] >= k)
            head++;
        while(tail >= head && f[i] > f[que[tail]])
            tail--;
        tail++;
        que[tail] = i;
        b[i][x] = f[que[head]];
    }
}
 
ll ans;
 
void getmaxb(int len){
    memset(que,0,sizeof que);
    int i,head = 1,tail = 1;
    for(i = 1;i < k;++i){
        while(tail >= head && f[i] > f[que[tail]])
            tail--;
        tail++;
        que[tail] = i;
    }
    for(i = k;i <= len;++i){
        while(tail >= head && i - que[head] >= k)
            head++;
        while(tail >= head && f[i] > f[que[tail]])
            tail--;
        tail++;
        que[tail] = i;
        ans += f[que[head]];
    }
}
 
int GCD(int x,int y){
    return !y?x:GCD(y,x%y);
}
 
int LCM(int x,int y){
    return x/GCD(x,y)*y;
}
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for(int i = 1;i <= n;++i)
        for(int j = 1;j <= m;++j)
            a[i][j] = LCM(i,j);
    for(int i = 1;i <= m;++i){
        for(int j = 1;j <= n;++j)
            f[j] = a[j][i];
        getmaxa(i,n);
    }
    for(int i = k;i <= n;++i){
        for(int j = 1;j <= m;++j)
            f[j] = b[i][j];
        getmaxb(m);
    }
    cout<<ans<<endl;
    return 0;
}