Palace
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 404 Accepted Submission(s): 104
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
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
题目大意：给出平面上n个点的坐标，可以得到平面最近点对的距离。
现求分别删掉第1~n个点时，平面上最近点对的距离的和。
对于不删点时可以求得最近点对。
如果删掉的点不是最近点对的两个端点，最近距离还是这两个点的距离。（会有n-2）次
其次，如果删除其中任何一个点，标记一下，然后再分别跑一边最近点，统计即可。
最近点对用分治法。
#include <iostream> #include <cmath> #include <vector> #include <cstdlib> #include <cstdio> #include <cstring> #include <queue> #include <stack> #include <list> #include <algorithm> #include <map> #include <set> #define LL long long #define Pr pair<int,int> #define fread(ch) freopen(ch,"r",stdin) #define fwrite(ch) freopen(ch,"w",stdout) using namespace std; const LL INF = 0x3f3f3f3f3f3f3f3f; const int msz = 100100; const int mod = 1e9+7; const double eps = 1e-8; struct Point { LL x,y; int id; }; LL dist(Point a,Point b) { return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); } Point p[msz]; Point tmpt[msz]; bool cmpxy(Point a,Point b) { if(a.x != b.x) return a.x < b.x; return a.y < b.y; } bool cmpy(Point a,Point b) { return a.y < b.y; } LL d; int pt[2]; void Closest_Pair(int left,int right,int opt) { if(left == right) return; LL tmp; if(left + 1 == right) { if(p[left].id != opt && p[right].id != opt) { tmp = dist(p[left],p[right]); if(d > tmp) { d = tmp; if(opt == -1) { pt[0] = p[left].id; pt[1] = p[right].id; } } } return; } int mid = (left+right)/2; Closest_Pair(left,mid,opt); Closest_Pair(mid+1,right,opt); int k = 0; for(int i = left; i <= right; ++i) { if(p[i].id != opt && abs(p[mid].x - p[i].x) <= d) tmpt[k++] = p[i]; } sort(tmpt,tmpt+k,cmpy); for(int i = 0; i < k; ++i) { for(int j = i+1; j < k && tmpt[j].y - tmpt[i].y < d; ++j) { tmp = dist(tmpt[i],tmpt[j]); if(d > tmp) { if(opt == -1) { pt[0] = tmpt[i].id; pt[1] = tmpt[j].id; } d = tmp; } } } } int main() { //fread(""); //fwrite(""); int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i = 0; i < n; ++i) { scanf("%lld%lld",&p[i].x,&p[i].y); p[i].id = i; } sort(p,p+n,cmpxy); d = INF; Closest_Pair(0,n-1,-1); LL ans = 1LL*d*(n-2); d = INF; Closest_Pair(0,n-1,pt[0]); ans += d; d = INF; Closest_Pair(0,n-1,pt[1]); ans += d; printf("%lld\n",ans); } return 0; }