P2874 [USACO07FEB]新牛棚Building A New Barn

题目描述

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? 给出平面上n个不相邻的点，要求到这n个点的曼哈顿距离之和最小的点的个数ans2，和这个最小距离ans1。

输入输出格式

输入格式：

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

输出格式：

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.

输入输出样例

输入样例#1： 

4
1 -3
0 1
-2 1
1 -1

输出样例#1： 

10 4

说明

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).

Solution:

　　一道数学+贪心的题目，思路正确，代码调了我一下午。

　　首先，我们知道两点曼哈顿距离公式：$d=|x1-x2|+|y1-y2|$

　　设$S$为题中所求的某个点$(x,y)$到$n$个点的曼哈顿距离之和，即$S=sum{d}$

　　那么有$S=sum{|x-xi|}+sum{|y-yi|}$

　　容易发现，$x,y$互不影响求值，于是，我们可以将本题中的横纵坐标分开各自计算。

　　这样本题就变成了一道中学数学题：求$|x-x1|+|x-x2|+…|x-xn|;min$，和求$|y-y1|+|y-y2|+…|y-yn|;min$

　　由中学的数学知识可知直接求中位数就行了（可以分类讨论，也可以几何意义解决）。

　　那么，我们直接对$x,y$各自排序：

　　 1、当$n$为奇数时，取$x[n/2+1]$和$y[n/2+1]$，由于题意中不能选给出的点，所以判断：

　　　　若该点为给出的点，则枚举它的上下左右四个点上能求得的最小的$S$，更新$ans1$，然后统计$ans$当且仅当这$4$个点的$S=ans1$;

　　　　若该点不为给出的点，则直接将$ans1$赋为$S$，$ans2$赋为$1$。

　　 2、当$n$为偶数时，取$x[n/2],x[n/2+1]$和$y[n/2],y[n/2+1]$，显然这$(x[n/2+1]-x[n/2]+1)*(y[n/2+1]-y[n/2]+1)$个点到给定的$n$个点的曼哈顿距离和$S$相等（因为这个矩形中的点的横坐标在$x[n/2],x[n/2+1]$之间，纵坐标也同理），于是枚举矩形中每个点是否是给定的点，求一次$ans1$就$OK$了，先让$ans2$等于这个矩形的点数，每次更新减小就行了。

代码：

#include<bits/stdc++.h>
#define il inline
#define ll long long
#define debug printf("%d %s
",__LINE__,__FUNCTION__)
using namespace std;
const int N=1e5+7;
int n,b[N],c[N],ans1=233333333,ans2;
struct point{int x,y;}a[N];
int dx[4]={1,-1,0,0},dy[4]={0,0,1,-1};
il int gi(){
    int a=0;char x=getchar();bool f=0;
    while((x<'0'||x>'9')&&x!='-')x=getchar();
    if(x=='-')x=getchar(),f=1;
    while(x>='0'&&x<='9')a=a*10+x-48,x=getchar();
    return f?-a:a;
}
il int getsum(int x,int y){
    int ans=0;
    for(int i=1;i<=n;i++)ans+=abs(x-b[i])+abs(y-c[i]);
    return ans;
}
int main()
{
    n=gi();
    for(int i=1;i<=n;i++)a[i].x=b[i]=gi(),a[i].y=c[i]=gi();
    sort(b+1,b+n+1);
    sort(c+1,c+n+1);
    if(n&1){
        int x=b[(n>>1)+1],y=c[(n>>1)+1];
        for(int i=0;i<4;i++){
            int xx=x+dx[i],yy=y+dy[i];
            int sum=getsum(xx,yy);
            if(sum<ans1)ans1=sum,ans2=1;
            else if(sum==ans1)++ans2;
        }
    }
    else {
        int x1=b[(n>>1)+1],x2=b[n>>1],y1=c[(n>>1)+1],y2=c[n>>1];
        ans2=(x1-x2+1)*(y1-y2+1);
        ans1=getsum(x1,y1);
        for(int i=1;i<=n;i++)
            if(a[i].x>=x2&&a[i].x<=x1&&a[i].y>=y2&&a[i].y<=y1)ans2--;
    }
    cout<<ans1<<' '<<ans2;
    return 0;
}