poj3714 Raid (divide and conquer find the nearest point on the plane)

Topic link: https: //vjudge.net/problem/POJ-3714

Meaning of the questions: Given two set points, for the shortest distance.

Ideas: the nearest point on the plane on the basis added a condition, what I do not visit marked with f, f 1 is a collection of 1, f 2 of the set to -1, then find the distance between two points, if af * calculating the distance bf = -1, or if the product is returned inf 1. And on the other hdoj1007 same.

AC Code:

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;

const int maxn=2e5+5;
const double inf=1e30;
int T,n,cnt,id[maxn];
struct node{
    int x,y,f;
}pt[maxn];

bool operator < (const node& a,const node& b){
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}

bool cmp(int a,int b){
    return pt[a].y<pt[b].y;
}

double dist(const node& a,const node& b){
    if(a.f*b.f==1) return inf;
    return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y));
}

double fenzhi(int l,int r){
    double d=inf;
    if(l==r) return d;
    if(l+1==r) return dist(pt[l],pt[r]);
    int mid=(l+r)>>1;
    d=min(fenzhi(l,mid),fenzhi(mid+1,r));
    cnt=0;
    int t1,t2,l1=l,r1=mid,mid1;
    while(l1<=r1){
        mid1=(l1+r1)>>1;
        if(pt[mid].x-pt[mid1].x<d) r1=mid1-1;
        else l1=mid1+1;
    }
    t1=l1;
    l1=mid+1,r1=r;
    while(l1<=r1){
        mid1=(l1+r1)>>1;
        if(pt[mid1].x-pt[mid].x<d) l1=mid1+1;
        else r1=mid1-1;
    }
    t2=r1;
    for(int i=t1;i<=t2;++i)
            id[++cnt]=i;
    sort(id+1,id+cnt+1,cmp);
    for(int i=1;i<cnt;++i)
        for(int j=i+1;j<=cnt&&(pt[id[j]].y-pt[id[i]].y<d);++j)
            d=min(d,dist(pt[id[i]],pt[id[j]]));
    return d;
}

int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=2*n;++i){
            scanf("%d%d",&pt[i].x,&pt[i].y);
            if(i<=n) pt[i].f=1;
            else pt[i].f=-1;
        }
        sort(pt+1,pt+1+2*n);
        printf("%.3f\n",fenzhi(1,2*n));
    }
    return 0;
}