POJ3714 Raid 平面最近点对

利用分治来求平面最近点对

只需要查后面6个点就好了

原因在于https://blog.csdn.net/liufeng_king/article/details/8484284

两个集合的话就把不同集合的设为inf就好了

const db inf = 1e100;

ll sqr(ll x) { return x*x; }

struct point {
    int x,y;
    int f;
    void in() {
        scanf("%d%d",&x,&y);
    }
    point operator-(const point &b) const {
        point ret;
        ret.x=x-b.x;
        ret.y=y-b.y;
        return ret;
    }
    db mean() {
        ll ret=0;
        ret=sqr(x)+sqr(y);
        return sqrt(ret);
    }
};

point a[N];

db dis(point a,point b) {
    db ret=0;
    if(a.f==b.f) return inf;
    ret=(a-b).mean();
    return ret;
}

bool cmp_x(point a,point b) {
    return a.x<b.x;
}

int t[N];
bool cmp_t(int x,int y) {
    return a[t[x]].y<a[t[y]].y;
}

db F(int l,int r) {
    if(l==r) return inf;
    if(l+1==r) return dis(a[r],a[l]);
    int mid=(l+r)>>1;
    db ans1=F(l,mid);
    db ans2=F(mid+1,r);
    db ans=min(ans1,ans2);
    if(ans==0) return ans;
    int nm=0;
    for(int i=l;i<=r;i++) if(a[mid].x-ans<=a[i].x && a[i].x<=a[mid].x+ans) {
        t[++nm]=i;
    }
    sort(t+1,t+nm+1,cmp_t);
    for(int i=1;i<=nm;i++) {
        for(int j=i+1;j<=min(nm,i+6);j++) {
            db anss=dis(a[t[i]],a[t[j]]);
            ans=min(ans,anss);
        }
    }
    return ans;
}

int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++) {
            a[i].in();
            a[i].f=1;
        }
        for(int i=n+1;i<=2*n;i++) {
            a[i].in();
            a[i].f=2;
        }
        sort(a+1,a+2*n+1,cmp_x);
        db ans=F(1,2*n);
        printf("%.3f
",ans);
    }
    return 0;
}