分治法的典例
当练手了
神奇的是,使用inplace_merge按说应该是O(n)的算法,但是用sort nlogn的算法反而更快
先上快排版
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> using namespace std; const int SIZE = 10000+10; const double INF = 100000; struct Point { double x,y; }p[SIZE],q[SIZE]; int n; inline double dis(const Point a, const Point b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } bool cmpx(const Point &a, const Point &b) { return a.x<b.x; } bool cmpy(const Point &a, const Point &b) { return a.y<b.y; } double MinDis(int l,int r) { if(r-l<1)return INF; if(r-l==1)return dis(p[r],p[l]); int m=(l+r)/2; double d=min(MinDis(l,m),MinDis(m+1,r)); int left=l,right=r+1; int i; for(i=m;i>=l;i--) if(p[i].x<p[m].x-d) { left=i+1; break; } for(i=m;i<=r;i++) if(p[i].x>p[m].x+d) { right=i; break; } for(int i=left;i<right;i++)q[i]=p[i]; sort(q+left,q+right,cmpy); int j; double ret=d; for(int i=left;i<right;i++) { for(j=i+1;j<right && q[j].y-q[i].y<d;j++) { ret=min(ret,dis(q[i],q[j])); } } return ret; } int main() { double ans; 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,cmpx); ans=MinDis(0,n-1); if(ans>10000)printf("INFINITY\n"); else printf("%.4lf\n",ans); } return 0; }
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> using namespace std; const int SIZE = 10000+10; const double INF = 100000; #define dis(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)) #define Fabs(x) x>0?x:-x const double eps=1e-9; struct Point { double x,y; }p[SIZE],q[SIZE]; int n; /*inline double dis(const Point a, const Point b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); }*/ bool cmpx(const Point &a, const Point &b) { return a.x<b.x; } bool cmpy(const Point &a, const Point &b) { return a.y<b.y; } double MinDis(int l,int r) { if(r-l<1)return INF; //if(r-l==1)return dis(p[r],p[l]); int m=(l+r)/2; double d=min(MinDis(l,m),MinDis(m+1,r)); inplace_merge(p+l,p+m+1,p+r+1,cmpy); int j,right=0; for(int i=l;i<=r;i++)//0 <n if(p[i].x >= p[m].x-d && p[i].x<=p[m].x+d) q[right++]=p[i]; double ret=INF; for(int i=0;i<right;i++) { for(j=i+1;j<right && q[j].y-q[i].y<d;j++) { ret=min(ret,dis(q[i],q[j])); } } return min(d,ret); } int main() { double ans; 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,cmpx); ans=MinDis(0,n-1); if(ans>=10000)printf("INFINITY\n"); else printf("%.4lf\n",ans); } return 0; }