After scrimping and saving for years, Farmer John has decided to build a new barn. He wants the barn to be highly accessible, and he knows the coordinates of the grazing spots of all N (2 ≤ N ≤ 10,000 cows. Each grazing spot is at a point with integer coordinates (Xi, Yi) (-10,000 ≤ Xi ≤ 10,000; -10,000 ≤ Yi ≤ 10,000). The hungry cows never graze in spots that are horizontally or vertically adjacent.
The barn must be placed at integer coordinates and cannot be on any cow's grazing spot. The inconvenience of the barn for any cow is given the Manhattan distance formula | X - Xi | + | Y - Yi|, where (X, Y) and (Xi, Yi) are the coordinates of the barn and the cow's grazing spot, respectively. Where should the barn be constructed in order to minimize the sum of its inconvenience for all the cows?
Input
Line 1: A single integer: N
Lines 2.. N+1: Line i+1 contains two space-separated integers which are the grazing location ( Xi, Yi) of cow i
Output
Line 1: Two space-separated integers: the minimum inconvenience for the barn and the number of spots on which Farmer John can build the barn to achieve this minimum.
Sample Input
4 1 -3 0 1 -2 1 1 -1
Sample Output
10 4
Hint
The minimum inconvenience is 10, and there are 4 spots that Farmer John can build the farm to achieve this: (0, -1), (0, 0), (1, 0), and (1, 1).
题意：给出n个点及其坐标，求一点到其它点曼哈顿距离总和最小值以及解的个数。
两点曼哈顿距离公式：d=|x1−x2|+|y1−y2|，由于两点曼哈顿距离的特性，单独求 x 与单独求 y 互不影响
因此，题目即为求 |x−x1|+|x−x2|+…+|x−xn| 的最小值，求 |y−y1|+|y−y2|+…+|y−yn| 的最小值，直接求两者中位数即可。
接对 x、y 各自进行排序比较：
1）当n为奇数时，取（ x[n/2+1]，y[n/2+1] ）
若该点为给出点，枚举它的上下左右四个方向上的点能求的最小的 d，然后统计当且仅当这4个点的方案数；
若该点不为给出点，则直接记录最小距离，方案数为1。
2）当n为偶数时，取（ x[n/2]，y[n/2] ）和（ x[n/2+1]，y[n/2+1] ）
由曼哈顿距离的特性知：共有（ x[n/2+1]−x[n/2]+1）*（ y[n/2+1]−y[n/2]+1）个点构成一个方块，且它们到给定的n个点（包括在方块内的点）的曼哈顿距离和d（所需求的距离）相等。由于题目说cows never graze in spots that are horizontally or vertically adjacent，（牛不会水平或者垂直相邻）那么被占满的情况就只有n是奇数的情况。因此枚举每个点是否为给定的点，求一次最小距离，再先让方案数为点的个数，每次更新减小方案数即可。
补充一个样例：
input：
4
1 1
2 2
3 3
4 4
output：
8 2
贴代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 10001
#define MOD 123
#define E 1e-6
using namespace std;
int dx[4]= {0,1,0,-1};
int dy[4]= {1,0,-1,0};
struct Node
{
int x;
int y;
} a[N];
int x[N],y[N];
int minn,plan;
int main()
{
int n;
scanf("%d",&n);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
x[i]=a[i].x;
y[i]=a[i].y;
}
sort(x+1,x+n+1);
sort(y+1,y+n+1);
if(n%2)
{
int temp=(n/2)+1;
for(int i=1; i<=n; i++)
{
if(a[i].x==x[temp]&&a[i].y==y[temp])
{
int Min=INF;
for(int l=0; l<4; l++)
{
int xx=x[temp]+dx[l];
int yy=y[temp]+dy[l];
int sum=0;
for(i=1; i<=n; i++)
sum+=abs(a[i].x-xx)+abs(a[i].y-yy);
if(sum<Min)
{
Min=sum;
plan=1;
}
else if (sum==Min)
plan++;
}
printf("%d %d\n",Min,plan);
return 0;
}
else
{
minn+=abs(a[i].x-x[temp])+abs(a[i].y-y[temp]);
plan=1;
}
}
printf("%d %d\n",minn,plan);
}
else
{
int temp1=n/2,temp2=n/2+1;
plan=(x[temp2]-x[temp1]+1)*(y[temp2]-y[temp1]+1);
for(int i=1; i<=n; i++)
{
minn+=abs(a[i].x-x[temp1])+abs(a[i].y-y[temp1]);
int x0=a[i].x,y0=a[i].y;
if (x[temp1]<=x0&&x0<=x[temp2]&&y[temp1]<=y0&&y0<=y[temp2])
plan--;
}
printf("%d %d\n",minn,plan);
}
return 0;
}