uva 10245 近期点对问题

分治法的典例

当练手了

奇妙的是。使用inplace_merge按说应该是O(n)的算法。可是用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;
}

再上归并

#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;
}