题目链接：
Palace
Time Limit: 8000/4000 MS (Java/Others)
Memory Limit: 262144/262144 K (Java/Others)
Problem Description
The last trial Venus imposes on Psyche is a quest to the underworld. She is to take a box and obtain in it a dose of the beauty of Prosperina, queen of the underworld.
There are n palaces in the underworld, which can be located on a 2-Dimension plane with (x,y) coordinates (where x,y are integers). Psyche would like to find the distance of the closest pair of two palaces. It is the password to enter the main palace.
However, the underworld is mysterious and changes all the time. At different times, exactly one of the n palaces disappears.
Psyche wonders what the distance of the closest pair of two palaces is after some palace has disappeared.
Print the sum of the distance after every single palace has disappeared.
To avoid floating point error, define the distance d between palace (x1,y1) and (x2,y2) as d=(x1−x2)2+(y1−y2)2.
There are n palaces in the underworld, which can be located on a 2-Dimension plane with (x,y) coordinates (where x,y are integers). Psyche would like to find the distance of the closest pair of two palaces. It is the password to enter the main palace.
However, the underworld is mysterious and changes all the time. At different times, exactly one of the n palaces disappears.
Psyche wonders what the distance of the closest pair of two palaces is after some palace has disappeared.
Print the sum of the distance after every single palace has disappeared.
To avoid floating point error, define the distance d between palace (x1,y1) and (x2,y2) as d=(x1−x2)2+(y1−y2)2.
Input
The first line of the input contains an integer T (1≤T≤5), which denotes the number of testcases.
For each testcase, the first line contains an integers n (3≤n≤105), which denotes the number of temples in this testcase.
The following n lines contains n pairs of integers, the i-th pair (x,y) (−105≤x,y≤105) denotes the position of the i-th palace.
For each testcase, the first line contains an integers n (3≤n≤105), which denotes the number of temples in this testcase.
The following n lines contains n pairs of integers, the i-th pair (x,y) (−105≤x,y≤105) denotes the position of the i-th palace.
Output
For each testcase, print an integer which denotes the sum of the distance after every single palace has disappeared.
Sample Input
1
3
0 0
1 1
2 2
Sample Output
12
题意：
给定n个点,每次删除一个点,问剩下的最近点对距离和是多少;
思路：
先找一遍最近点对的两个点,删除其他n-2个点对最近点对没有影响,然后再分别删除最近点对的这两个点,再最近点对的距离,最后把这些加起来就好了;
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}
const LL mod=998244353;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e5+10;
const int maxn=1e3+10;
const double eps=1e-6;
struct node
{
LL x,y;
}po[N],temp[N],po1[N];
int cmp1(node a,node b)
{
return a.x<b.x;
}
int cmp2(node a,node b)
{
return a.y<b.y;
}
node fa,fb;
LL ans;
LL get_dis(node a, node b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
LL merge(int l,int r)
{
if(r-l<=1)
{
if(l==r)return inf;
else
{
LL dis=get_dis(po[l],po[r]);
if(dis<ans)
{
fa=po[l];
fb=po[r];
ans=dis;
}
return dis;
}
}
int mid=(l+r)>>1;
LL dl=merge(l,mid),dr=merge(mid+1,r),d=min(dl,dr);
int cnt=0;
For(i,l,r)
{
if(abs(po[mid].x-po[i].x)<=d)temp[++cnt]=po[i];
}
sort(temp+1,temp+cnt+1,cmp2);
For(i,1,cnt)
{
for(int j=i+1;j<=cnt&&temp[j].y-temp[i].y<d;j++)
{
LL dis=get_dis(temp[j],temp[i]);
if(dis<d)d=dis;
if(dis<ans)
{
ans=dis;
fa=temp[j];
fb=temp[i];
}
}
}
return d;
}
LL work(int l,int r)
{
ans=inf;
sort(po+l,po+r+1,cmp1);
return merge(l,r);
}
int main()
{
int t;
read(t);
while(t--)
{
int n;
read(n);
For(i,1,n)read(po1[i].x),read(po1[i].y),po[i]=po1[i];
LL sum=0;
sum+=work(1,n)*(LL)(n-2);
node faa=fa,fbb=fb;
int flag=0;
For(i,1,n)
{
if(po1[i].x==faa.x&&po1[i].y==faa.y&&flag==0)
{
flag=1;
po[i].x=1e7;
po[i].y=1e7;
continue;
}
po[i]=po1[i];
}
sum+=work(1,n);
flag=0;
For(i,1,n)
{
if(po1[i].x==fbb.x&&po1[i].y==fbb.y&&flag==0)
{
flag=1;
po[i].x=1e7;
po[i].y=1e7;
continue;
}
po[i]=po1[i];
}
sum+=work(1,n);
cout<<sum<<"
";
}
return 0;
}