poj 3714

递归分治法求最近点对。介绍个讲最近点对的地方。http://wenku.baidu.com/view/2baa55d349649b6648d74736.html 

这里粘一下小媛的代码，我加了一点注释，作为最近点对的模板。

/*需要先对x排序，然后分治的时候对y排序才对。精度判断WA了多次，最后找到，getmin如果改成小于等于就过了，或者不加精度判断也行，小于的话死活过不去，我把eps开到15次方才过><。。。郁闷*/#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;const int MAX = 100010;const double D_MAX = 1e50;const double eps = 1e-15;struct point{double x,y;};point p[MAX];bool dy(double x,double y)  // x > y {return x > y + eps;}bool xy(double x,double y) // x < y {return x < y - eps;}bool dyd(double x,double y) // x >= y {return x > y - eps;}bool xyd(double x,double y) // x <= y {return x < y + eps;}bool dd(double x,double y)  // x == y {return fabs( x - y ) < eps;}double disp2p(point a,point b){return sqrt( (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) );}double getmin(double a,double b){return xy(a,b) ? a : b;}bool cmp(point a,point b){if( dd(a.x,b.x) )return xy(a.y,b.y);return xy(a.x,b.x);}bool cmpy(point a,point b){if( dd(a.y,b.y) )return xy(a.x,b.x);return xy(a.y,b.y);}point y[MAX];double nearpot(int l,int r){if( l == r )return D_MAX;if( l == r-1 )return disp2p(p[l],p[r]);if( l == r-2 )return getmin( getmin(disp2p(p[l],p[l+1]),disp2p(p[l],p[l+2])), disp2p(p[l+1],p[l+2]) );int mid = (l+r)>>1;//对左右两边分别递归求解 最近点对。两者中取较小值。 double ans = getmin( nearpot(l,mid), nearpot(mid+1,r) );int len = 0;//将距中位线l=m的距离小于dist且宽度为2*dist的点置于z[]中for(int i=mid; i>=l && xy(p[mid].x-p[i].x,ans); i--)y[len++] = p[i];for(int i=mid+1; i<=r && xy(p[i].x-p[mid].x,ans); i++)y[len++] = p[i];sort(y,y+len,cmpy);//对Y排序 //分治法求最近点对 for(int i=0; i<len-1; i++)for(int k=i+1; k<len; k++){if( dyd(y[k].y-y[i].y,ans) )break;ans = getmin(ans,disp2p(y[i],y[k]));}return ans;}int main(){int n;while( ~scanf("%d",&n) && n ){for(int i=0; i<n; i++)scanf("%lf %lf",&p[i].x,&p[i].y);sort(p,p+n,cmp);//对X进行排序 double ans = nearpot(0,n-1);printf("%.2lf\n",ans/2.0);}return 0;}