Description
Under the guidance of the flower teacher, there is a gathering every 4 days, commonly known as "water" activities.
See BZOJ3153 for details of water activities
But that's not the point
There are many fantasy questions, each with two reference coefficients: code difficulty and algorithm difficulty.
Flower gods must find a topic that suits everyone as best as possible in order to prepare for a flower-watering party
Now Flower knows everyone's code ability X and algorithm ability y. One question (code difficulty X algorithm difficulty Y) is Max (abs (X - x)
, abs ( Y – y ) )
That is to say, whether it is too difficult or too simple, it will make the title unsuitable for you to do (if you set it all according to the capabilities of the goddess themselves, the rhythm of the absolute all-round explosion is too simple, it will not show the power of flourishing).
Of course, not every time, there is not necessarily a question for everyone, so make the sum of all people's inadequacies as low as possible
Flower God created 100001*100001 questions, each of which had code and algorithm difficulty of 0, 1, 2, 3,..., 100,000
Input
The first line has a positive integer N, indicating that there are N students in the flower god. The Flower God wants to choose a question for N students.
Next N lines, an integer x[i],y[i] separated by two spaces on each line, indicate the student's code and algorithm capabilities
Output
An integer representing the smallest improper moderate sum
Sample Input
3
1 2
2 1
3 3
Sample Output
3
HINT
For 100% of the data, n<=100000,0<=x[i], y[i]<=100000
Problem
I thought for a while last night and didn't come up with it.
Come this morning
suddenly
Is this not the Chebyshev distance
Turn Manhattan Distance Strongly
Scan once for median
The example tells me with good conscience that Chebyshev may not be the whole point...
Then you'll sweep the four dots next to him all over.
Why is this data so watery?
Should the answer not be less than 100,000?
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*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
LL ans;
int A[100005],B[100005];
int n;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
int x,y;scanf("%d%d",&x,&y);
A[i]=x+y;B[i]=x-y;
}
sort(A+1,A+1+n);sort(B+1,B+1+n);
int md1=(n+1)/2,md2=(n+1)/2;
if((A[md1]+B[md2])%2==0)
{
for(int i=1;i<=n;i++)ans+=abs(A[md1]-A[i])+abs(B[md2]-B[i]);
printf("%lld\n",ans/2);
}
else
{
int xpos=A[md1]-1,ypos=B[md2];LL hh=0;
for(int i=1;i<=n;i++)hh+=abs(xpos-A[i])+abs(ypos-B[i]);
ans=hh;
xpos=A[md1]+1;ypos=B[md2];hh=0;
for(int i=1;i<=n;i++)hh+=abs(xpos-A[i])+abs(ypos-B[i]);
ans=min(ans,hh);
xpos=A[md1];ypos=B[md2]-1;hh=0;
for(int i=1;i<=n;i++)hh+=abs(xpos-A[i])+abs(ypos-B[i]);
ans=min(ans,hh);
xpos=A[md1];ypos=B[md2]+1;hh=0;
for(int i=1;i<=n;i++)hh+=abs(xpos-A[i])+abs(ypos-B[i]);
ans=min(ans,hh);
printf("%lld\n",ans/2);
}
return 0;
}