http://acm.hdu.edu.cn/showproblem.php?pid=4311

http://acm.hdu.edu.cn/showproblem.php?pid=4312

4311 Finds the Sum of Manhattan Distances in Minimum Space and 4312 Finds the Sum of Chebyshev Distances in Minimum Space

Summary of Three Spatial Distances http://www.cnblogs.com/adelalove/p/8612540.html

Manhattan Distance

In Manhattan distance, x and y don't want to do with each other and sort separately, then find a prefix and suffix.

If the answer does not require a point, rank each dimension directly in the median.

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll N=0x3f3f3f3f3f3f3f3f;

struct node
{
    ll x,y;
    int id;
};

node point[100010];
ll pre[100010],post[100010];
int n;

bool cmpI(node n1,node n2)
{
    return n1.x<n2.x;
}

bool cmpII(node n1,node n2)
{
    return n1.y<n2.y;
}

ll getmin(ll a,ll b)
{
    if(a<b) return a;
    else return b;
}

int main()
{
    ll i,sum,ans;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            point[i].id=i;
            scanf("%lld%lld",&point[i].x,&point[i].y);
        }
        memset(pre,0,sizeof(pre));
        memset(post,0,sizeof(post));

        sort(point+1,point+n+1,cmpI);
        sum=0;
        for(i=2;i<=n;i++)
        {
            sum+=(i-1)*(point[i].x-point[i-1].x);
            pre[point[i].id]+=sum;
        }
        sum=0;
        for(i=n-1;i>=1;i--)
        {
            sum+=(n-i)*(point[i+1].x-point[i].x);
            post[point[i].id]+=sum;
        }

        sort(point+1,point+n+1,cmpII);
        sum=0;
        for(i=2;i<=n;i++)
        {
            sum+=(i-1)*(point[i].y-point[i-1].y);
            pre[point[i].id]+=sum;
        }
        sum=0;
        for(i=n-1;i>=1;i--)
        {
            sum+=(n-i)*(point[i+1].y-point[i].y);
            post[point[i].id]+=sum;
        }

        ans=N;
        for(i=1;i<=n;i++) ans=getmin(ans,pre[i]+post[i]);
        printf("%lld\n",ans);
    }
    return 0;
}

Chebyshev Distance

Unexpectedly, math is too bad and there are too few problems to solve.

For two points (x1,y1) (x2,y2)

dis = max(x2-x1,y2-y1) = ( |(x2-x1)+(y2-y1)| + |(x2-x1)-(y2-y1)| ) / 2 = ( |(x2+y2)-(x1+y1)| + |(x2-y2)-(x1-y1)| ) / 2

In this way, the Chebyshev distance between (x1,y1) (x2,y2) is converted to the Manhattan distance between (x1+y1,x1-y1) (x2+y2,x2-y2).

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll N=0x3f3f3f3f3f3f3f3f;

struct node
{
    ll x,y;
    int id;
};

node point[100010];
ll pre[100010],post[100010];
int n;

bool cmpI(node n1,node n2)
{
    return n1.x<n2.x;
}

bool cmpII(node n1,node n2)
{
    return n1.y<n2.y;
}

ll getmin(ll a,ll b)
{
    if(a<b) return a;
    else return b;
}

int main()
{
    ll i,x,y,sum,ans;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            point[i].id=i;
            scanf("%lld%lld",&x,&y);
            point[i].x=x+y,point[i].y=x-y;
        }
        memset(pre,0,sizeof(pre));
        memset(post,0,sizeof(post));

        sort(point+1,point+n+1,cmpI);
        sum=0;
        for(i=2;i<=n;i++)
        {
            sum+=(i-1)*(point[i].x-point[i-1].x);
            pre[point[i].id]+=sum;
        }
        sum=0;
        for(i=n-1;i>=1;i--)
        {
            sum+=(n-i)*(point[i+1].x-point[i].x);
            post[point[i].id]+=sum;
        }

        sort(point+1,point+n+1,cmpII);
        sum=0;
        for(i=2;i<=n;i++)
        {
            sum+=(i-1)*(point[i].y-point[i-1].y);
            pre[point[i].id]+=sum;
        }
        sum=0;
        for(i=n-1;i>=1;i--)
        {
            sum+=(n-i)*(point[i+1].y-point[i].y);
            post[point[i].id]+=sum;
        }

        ans=N;
        for(i=1;i<=n;i++) ans=getmin(ans,pre[i]+post[i]);
        printf("%lld\n",ans/2);
    }
    return 0;
}