PIPIOJ 1018: Soldiers line up greedy mathematics

http://pipioj.online/problem.php?id=1018

Idea: Very similar to this question . Since it is known that when $x=a[mid]$时，${\sum }_{i=1}^n|a_i-x|\; takes\; the$ minimum value, then the ordinate of moving all soldiers to the same row is determined. Now consider the abscissa, assuming that the abscissa of the first soldier in the final result is$t$ , then according to the greedy strategy, we must hope that the smallest$x_i$Move to $t$ , the second smallest$x_i$Move to $t+1$ ……Then after sorting the abscissa, we hope to find:
$$min(\sum _{i=1}^n|x_i-(t+i-1)|)=min(\sum _{i=1}^n|(x_i-i+1)-t|)$$
令$x_i^{\prime }=x_i-i+1$ , the above equation can be changed:$min({\sum }_{i=1}^n|x_i^{\prime }-t|)$ , obviously when$t=x^{\prime }\left[Mid\right]$is the minimum value.

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=1e4+5;

int n,x[maxn],y[maxn];

int main()
{
    
    
    while(~scanf("%d",&n))
    {
    
    
        for(int i=1;i<=n;i++)
            scanf("%d %d",&x[i],&y[i]);
        sort(x+1,x+1+n);
        for(int i=1;i<=n;i++)
            x[i]-=i-1;
        sort(x+1,x+1+n);
        sort(y+1,y+1+n);
        int xmid=x[(n+1)>>1];
        int ymid=y[(n+1)>>1];
        int cnt=0;
        for(int i=1;i<=n;i++)
            cnt+=abs(x[i]-xmid)+abs(y[i]-ymid);
        printf("%d\n",cnt);
    }
    return 0;
}