题目连接:Fake Maxpooling
题意:
给你n*m的矩阵,矩阵每个值 ,问n*m矩阵下,所有k*k子矩阵里的最大值之和是多少?
题解:
这题一开始以为是数论,找规律啥的,最后才发现就是个单调队列,先维护行这一维的最大值,会形成一个矩阵,然后再维护列这一维的最大值,最后就得出了所有子矩阵最大值之和。
代码:
手写队列:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
//extern "C"{void *__dso_handle=0;}
typedef long long ll;
typedef long double ld;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int N=5005;
vector<int> v[N];
struct Node
{
int num,val;
Node(): num(),val(){}
Node(int num,int val): num(num),val(val){}
};
int main()
{
int n,m,k;
cin >> n >> m >> k;
for(int i=1;i<=n;i++)
{
deque<pair<int, int> > q;
Node que[N];
int l=1,r=0;
for(int j=1;j<=m;j++)
{
int lcm=i/__gcd(i,j)*j;
while(l<=r && lcm>que[r].val)
r--;
que[++r]=Node(j,lcm);
while(l<=r && j-que[l].num>=k)
l++;
if(j>=k) v[i].pb(que[l].val);
}
}
ll ans=0;
for(int i=0;i<m-k+1;i++)
{
Node que[N];
int l=1,r=0;
for(int j=1;j<=n;j++)
{
while(l<=r && v[j][i]>que[r].val)
r--;
que[++r]=Node(j,v[j][i]);
while(l<=r && j-que[l].num>=k)
l++;
if(j>=k) ans+=que[l].val;
}
}
cout << ans << endl;
}
用STL库的deque
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
//extern "C"{void *__dso_handle=0;}
typedef long long ll;
typedef long double ld;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int N=5005;
vector<int> v[N];
int main()
{
int n,m,k;
cin >> n >> m >> k;
for(int i=1;i<=n;i++)
{
deque<pair<int, int> > q;
for(int j=1;j<=m;j++)
{
int lcm=i/__gcd(i,j)*j;
while(!q.empty() && lcm>q.back().second)
q.pop_back();
q.push_back(mp(j,lcm));
while(!q.empty() && j-q.front().first>=k)
q.pop_front();
if(j>=k) v[i].pb(q.front().second);
}
}
ll ans=0;
for(int i=0;i<m-k+1;i++)
{
deque<pii> q;
for(int j=1;j<=n;j++)
{
while(!q.empty() && v[j][i]>q.back().second)
q.pop_back();
q.push_back(mp(j,v[j][i]));
while(!q.empty() && j-q.front().first>=k)
q.pop_front();
if(j>=k) ans+=q.front().second;
}
}
cout << ans << endl;
}