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题目描述 Description |
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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 Assume that all |
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输入描述 Input Description |
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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. |
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输出描述 Output Description |
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For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places. |
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样例输入 Sample Input |
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2 0 0 1 1 2 1 1 1 1 3 -1.5 0 0 0 0 1.5 0 |
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样例输出 Sample Output |
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0.71 0.00 0.75 |
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数据范围及提示 Data Size & Hint |
之前的一些废话：是时候准备会考了。。
题解：求最近点对。首先把平面划分成两个部分，递归求出两个部分的答案为ans，然后，把离分割线距离小于ans/2的点一左一右算距离更新ans.
代码：
#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;
}
总结：