HDU 5721 Palace BestCoder 2nd Anniversary （平面最近点对）

Palace

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 465    Accepted Submission(s): 118

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 then 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 integerT(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.

Output

For each testcase, print an integer which denotes the sum of the distance after every single palace has disappeared.

Sample Input

130 01 12 2

Sample Output

12
Hint
If palace $ (0,0) $ disappears，$ d = (1-2) ^ 2 + (1 - 2) ^ 2 = 2 $;If palace $ (1,1) $ disappears，$ d = (0-2) ^ 2 + (0 - 2) ^ 2 = 8 $;If palace $ (2,2) $ disappears，$ d = (0-1) ^ 2 + (0-1) ^ 2 = 2 $;Thus the answer is $ 2 + 8 + 2 = 12 $。
 

Source

BestCoder 2nd Anniversary

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题意：

给出平面上n个点的坐标，可以求平面最近点对的距离。

要求分别删掉第1~n个点时，平面上最近点对的距离的和。（防止精度问题，题目要求是距离的平方）

题解：求平面最近点对。分治法。

AC代码：

#include<iostream>#include<stdio.h>#include<cstring>#include<algorithm>typedef long long LL;using namespace std;const LL Max =1LL<<60;//平面最近点对:分治struct point{LL x,y;};point p[100010]; //数组开小会TLE int t[1000010];  //数组开小会TLE LL min(LL a,LL b){return a>b?b:a;}int cmp (const point &a,const point &b) //对点的x坐标排序 {if(a.x==b.x)return a.y<b.y;return a.x>b.x;}int cmpy(const int &a,const int &b) //对点的y坐标排序 {return p[a].y<p[b].y;} LL dis(point a,point b)   //两点的距离 {return ((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); }  LL myabs(LL x){return x<0?-x:x; } struct node {LL res;int x,y;node (){}node (LL res,int x,int y):res(res),x(x),y(y){}};LL sqr(LL x){return x*x; }  node solve(int left,int right)//分治，递归  { LL x=Max;  if(left==right)return node(x,0,0); if(left==right+1)return node(dis(p[left],p[right]),left,right); int mid=(left+right)>>1; //求出中间点  node dl=solve(left,mid); node dr=solve(mid+1,right); if(dl.res>dr.res) dl=dr; //把中间2dl宽度的部分的点作单独处理  int i,j,k; k=0; for(i=left;i<=right;i++)  { if(sqr(p[i].x-p[i+1].x)<dl.res) t[k++]=i; } sort(t,t+k,cmpy); //由下到上排序，扫描 for(i=0;i<k;i++)  { for(j=i+1;j<k&&sqr(p[t[j]].y-p[t[i]].y)<dl.res;j++) { LL d3 =dis(p[t[i]],p[t[j]]); if(dl.res>d3) { dl =node(d3,t[i],t[j]); } } } return dl; } int main() { int T; cin>>T; int n; while(T--) { cin>>n; int x,y; for(int i=1;i<=n;i++) { scanf("%d%d",&x,&y); p[i].x =x; p[i].y =y; }sort(p+1,p+n+1,cmp);node r=solve(1,n);LL res =r.res*(n-2);point tmp =p[r.x];p[r.x].x=1e9;p[r.x].y=1e9;res+=solve(1,n).res;p[r.x]=tmp;p[r.y].x=1e9;res+=solve(1,n).res;cout<<res<<endl; } return 0; }