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最新推荐文章于 2025-08-15 21:17:00 发布

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#基础题

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本文介绍了一种通过独立最小化矩阵中特定表达式的算法，该算法能够有效地找到使矩阵方差达到最小值的参数x和y。文章详细阐述了算法的原理，并提供了一个具体的实现示例。

刚开始根本读不懂题目。。。

官方题解

We need to find such x and y that the value of $\text{[math]}$ is minimum possible. This expression can be rewritten as $\text{[math]}$ . Note that the first part doesn't depend on y and the second part doesn't depend on x, so we can minimize these parts separately. Here is how to minimize $\text{[math]}$ , the second part is minimized similarly. As the expression in the brackets doesn't depend on j, this part can be rewritten as $\text{[math]}$ , where $\text{[math]}$ . Now it's enough to calculate the required value for all possible values of x and choose x for which this value is the smallest. The optimal value of y can be found similarly.

The overall complexity of this solution is O(n·m + n² + m²).

As the objective function is convex, other approaches to this problem are possible, for example, ternary search, gradient descent or analytical approach (calculation of derivatives).

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
#define N 1010
const long long inf=1LL<<60;
int n,m,a[N][N];
long long dx[N],dy[N];
int main()
{
    scanf("%d%d",&n,&m);
    memset(dx,0,sizeof(dx));
    memset(dy,0,sizeof(dy));
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            scanf("%d",&a[i][j]);
            dx[i]+=a[i][j];
            dy[j]+=a[i][j];
        }
    }
    long long mx=inf,my=inf;
    int nx,ny;
    for(int i=0;i<=n;i++)
    {
        long long ans=0;
        for(int j=0;j<n;j++)
        {
           ans+=(4*i-4*j-2)*(4*i-4*j-2)*dx[j];
        }
        if(ans<mx)
        {mx=ans;nx=i;}
    }
    for(int i=0;i<=m;i++)
    {
        long long ans=0;
        for(int j=0;j<m;j++)
        {
            ans+=(4*i-4*j-2)*(4*i-4*j-2)*dy[j];
        }
        if(ans<my)
        {my=ans;ny=i;}
    }
    printf("%I64d\n",mx+my);
    printf("%d %d\n",nx,ny);
    return 0;
}