[HDU1007]Quoit Design
|
題目描述 Description |
|
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. |
|
輸入描述 Input Description |
|
The 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. |
|
輸出描述 Output Description |
|
For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places. |
|
樣例輸入 Sample Input |
|
2 0 0 1 1 2 1 1 1 1 3 -1.5 0 0 0 0 1.5 0 |
|
樣例輸出 Sample Output |
|
0.71 0.00 0.75 |
|
數據範圍及提示 Data Size & Hint |
之前的一些廢話：是時候準備會考了。。
題解：求最近點對。首先把平面劃分成兩個部分，遞歸求出兩個部分的答案為ans，然後，把離分割線距離小於ans/2的點一左一右算距離更新ans.
代碼：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> using namespace std; typedef long long LL; #define mem(a,b) memset(a,b,sizeof(a)) typedef pair<int,int> PII; const int maxn=100010; const double oo=2147483647; struct Point { double x,y; Point() {} Point(double _1,double _2):x(_1),y(_2){} bool operator < (const Point &s)const { if(x==s.x)return y<s.y; return x<s.x; } }p[maxn]; int n;double x,y; double dis(double a,double b,double c,double d){ return sqrt((c-a)*(c-a)+(d-b)*(d-b)); } double mdis(int l,int r) { if(l==r)return 0; if(l+1==r)return dis(p[l].x,p[l].y,p[r].x,p[r].y); double ret=oo; int mid=(l+r)>>1; ret=min(mdis(l,mid),mdis(mid,r)); for(int i=mid-1;i>=l && p[mid].x-p[i].x<ret;i--) for(int j=mid+1;j<=r && p[j].x-p[mid].x<ret && fabs(p[i].y-p[j].y)<ret;j++) ret=min(ret,dis(p[i].x,p[i].y,p[j].x,p[j].y)); return ret; } int main() { while(scanf("%d",&n)!=EOF && n) { for(int i=0;i<n;i++)scanf("%lf%lf",&x,&y),p[i]=Point(x,y); sort(p,p+n); printf("%.2lf\n",mdis(0,n-1)/2); for(int i=0;i<n;i++)p[i]=Point(0,0); } return 0; }View Code
總結：
[HDU1007]Quoit Design