HDU - 1007 平面最近点对

Posted 2020-10-31 hit_yjl

Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded. 
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring. 

Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0. 

InputThe input consists of several test cases. For each case, the first line contains an integer N (2 <= N <= 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0. 
OutputFor each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places. 
Sample Input

2
0 0
1 1
2
1 1
1 1
3
-1.5 0
0 0
0 1.5
0

Sample Output

0.71
0.00
0.75

思路：每次按x坐标分成左右两坨，处理完左右两坨后考虑中间相邻的那部分，考虑可能更新答案的点集，按y值双指针移动即可，一个trick是合并两个子问题时同时类似归并排序将点按y值顺便排好序而不是每次快排重新排序。
每次合并代价O（n），这样最终复杂度nlogn。

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 struct point {
 45     double x, y;
 46     void input() {
 47         scanf("%lf%lf", &x, &y);
 48     }
 49     double dis(const point &p) {
 50         return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
 51     }
 52     bool operator < (const point &p) const {
 53         return x < p.x;}
 54 } p[100015], q[100015], p1[100015], p2[100015];
 55 int n;
 56 double dis;
 57 void mymerge(int s, int e) {
 58     if (s == e) return;
 59     int mi = s + e >> 1;
 60     mymerge(s, mi);
 61     mymerge(mi + 1, e);
 62     double d2 = p[mi].x + dis;
 63     double d1 = p[mi + 1].x - dis;
 64     int i = s, j = mi + 1, k = s, i1 = 1, i2 = 1;
 65     while (i <= mi && j <= e) {
 66         if (p[i].y < p[j].y) {
 67             if (d1 - EPS <= p[i].x && p[i].x <= d2 + EPS) {
 68                 p1[i1++] = p[i];
 69             }
 70             q[k++] = p[i++];
 71         }
 72         else {
 73             if (d1 - EPS <= p[j].x && p[j].x <= d2 + EPS) {
 74                 p2[i2++] = p[j];
 75             }
 76             q[k++] = p[j++];
 77         }
 78     }
 79     while (i <= mi) {
 80         if (d1 - EPS <= p[i].x && p[i].x <= d2 + EPS) {
 81             p1[i1++] = p[i];
 82         }
 83         q[k++] = p[i++];
 84     }
 85     while (j <= e) {
 86         if (d1 - EPS <= p[j].x && p[j].x <= d2 + EPS) {
 87             p2[i2++] = p[j];
 88         }
 89         q[k++] = p[j++];
 90     }
 91     for (int i = s; i <= e; ++i) p[i] = q[i];
 92     for (int i = 1, j = 1; i < i1 && j < i2; ++i) {
 93         while (j < i2 && p2[j].y < p1[i].y - dis) ++j;
 94         for (int k = j; k < i2 && p2[j].y < p1[i].y + dis; ++k) {
 95             double td = p1[i].dis(p2[j]);
 96             dis = min(dis, td);
 97         }
 98     }
 99 }
100 int main() {
101     while (~scanf("%d", &n) && n) {
102         for (int i = 1; i <= n; ++i) {
103             p[i].input();
104         }
105         sort(p + 1, p + n + 1);
106         dis = 1e18;
107         mymerge(1, n);
108         printf("%.2f\n", dis / 2);
109     }
110     return 0;
111 }

View Code