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Description
在花老师的指导下,每周4都有一个集会活动,俗称“浇水”活动。
具体浇水活动详情请见BZOJ3153
但这不是重点
花神出了好多题,每道题都有两个参考系数:代码难度和算法难度
花神为了准备浇花集会的题,必须找一道尽量适合所有人的题
现在花神知道每个人的代码能力x和算法能力y,一道题(代码难度X算法难度Y)对这个人的不适合度为 Max ( abs ( X – x )
, abs ( Y – y ) )也就是说无论太难还是太简单都会导致题目不适合做(如果全按花神本人能力设题,绝对的全场爆0的节奏,太简单,则体现不出花神的实力)
当然不是每次都如花神所愿,不一定有一道题适合所有人,所以要使所有人的不合适度总和尽可能低
花神出了100001*100001道题,每道题的代码难度和算法难度都为0,1,2,3,……,100000
Input
第一行一个正整数N,表示花神有N个学生,花神要为这N个学生选一道题
接下来N行,每行两个空格隔开的整数x[i],y[i],表示这个学生的代码能力和算法能力
Output
一个整数,表示最小的不合适度总和
Sample Input
3
1 2
2 1
3 3
Sample Output
3
HINT
对于100%的数据,n<=100000,0<=x[i],y[i]<=100000
题解
昨晚想了一会没想出来…
今早过来
突然
这不是切比雪夫距离吗
大力转曼哈顿距离
求中位数扫一遍
样例很良心告诉我切比雪夫可能不是整点…
那你就把他旁边的四个点扫一遍啊…
这数据怎么这么水啊
不是应该还判答案是否小于100000吗
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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;
}