hdu 3124 Moonmist

http://acm.hdu.edu.cn/showproblem.php?pid=3124

题意：给出n个相离的圆，求最近的不同圆上两点的距离

二分答案a

所有圆的半径增加a，若此时有圆相交，说明最近距离小于a

否则，最近距离大于a

如何判断是否有圆相交？

扫描线从左往右扫，用set维护此时不相交的圆，按圆心的纵坐标升序排列

扫到一个圆的最左端，判断该圆加入set会不会与set里的圆相交

因为set里的圆都已经按圆心纵坐标升序排列，只判断该圆圆心上下的两个圆

所以扫到一个圆的最右端，删除圆的时候，也要判断该圆的上下两个圆是否有交点

因为一个圆可能隔着它与另一个相交

#include<set>
#include<cmath>
#include<cstdio>
#include<algorithm>

using namespace std;

#define N 50001

int n;
double MID,mid;
struct Circle
{
    int x,y,r;
}cir[N];

struct Line
{
    int id;
    double xi;
}e[N<<1]; 

struct node
{
    int id;
    
    node(int id_) : id(id_) {}
    
    bool operator < (const node p) const
    {
        return id<p.id;
    }
};
set<node>s;
set<node>::iterator it,itt;

bool cmp(Circle p,Circle q)
{
    if(p.y!=q.y) return p.y<q.y;
    return p.x<q.x;
} 

bool cmp2(Line p,Line q)
{
    return p.xi<q.xi;
}

bool point(int i,int j)
{
    double dis=sqrt(1.0*(cir[i].x-cir[j].x)*(cir[i].x-cir[j].x)+1.0*(cir[i].y-cir[j].y)*(cir[i].y-cir[j].y));
    return cir[i].r+cir[j].r+2*mid>=dis;
}

bool in(int d)
{
    if(s.empty())
    {
        s.insert(node(d));
        return true;
    } 
    it=s.upper_bound(node(d));
    if(it!=s.end()) 
        if(point((*it).id%n,d)) return false;
    if(it==s.begin()) 
    {
        s.insert(node(d));
        return true;
    }
    it--;
    if(point((*it).id%n,d)) return false;
    s.insert(node(d));
    return true;
}

bool out(int d)
{
    s.erase(node(d));
    it=s.upper_bound(node(d));
    if(it==s.end()) return true;
    if(it==s.begin()) return true;
    itt=it;
    itt--;
    if(point((*it).id%n,(*itt).id%n)) return false;
    return true;
}        

bool check()
{
//    printf("%.6lf
",MID); 
    s.clear();
    mid=MID/2;
    int m=0;
    for(int i=0;i<n;++i)
    {
        e[++m].id=i;
        e[m].xi=cir[i].x-cir[i].r-mid;
        e[++m].id=i+n;
        e[m].xi=cir[i].x+cir[i].r+mid;
    }
    sort(e+1,e+m+1,cmp2);
    int k;
    for(int i=1;i<=m;++i)
    {
        k=e[i].id%n;
        if(e[i].id<n)
        {
            if(!in(k)) return false;
        }
        else
        {
            if(!out(k)) return false;
        }
    /*    it=s.begin();
        while(it!=s.end())
        {
            printf("%d",(*it).id%n);
            it++;
        }
        printf("
");*/
    }
    return true;
}
            
int main()
{
//    freopen("data.txt","r",stdin);
    //freopen("my.txt","w",stdout); 
    int T;
    scanf("%d",&T);
    double l,r;
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;++i) scanf("%d%d%d",&cir[i].x,&cir[i].y,&cir[i].r);
        sort(cir,cir+n,cmp);
        l=0,r=1000000;
        while(r-l>1e-10)
        {
            MID=(l+r)/2;
            if(check()) l=MID;
            else r=MID;
        }
        printf("%.6lf
",MID);
    }
    return 0;
}

Moonmist

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2293    Accepted Submission(s): 820

Problem Description

An Unidentified Flying Object (Commonly abbreviated as UFO) is the popular term for any aerial phenomenon whose cause cannot be easily or immediately identified. We always believe UFO is the vehicle of aliens. But there is an interrogation about why UFO always likes a circular saucer? There must be a reason. Actually our scientists are developing a new traffic system “Moonmist”. It is distinguished from the traditional traffic. We use circular saucers in this new traffic system and the saucers moves extremely fast. When our scientists did their test, they found that traffic accident was too hard to be avoided because of the high speed of the advanced saucer. They need us to develop a system that can tell them the nearest saucer. The distance between two saucers is defined as the shortest distance between any two points in different saucers.

 

Input

The first line consists of an integer T, indicating the number of test cases.
The first line of each case consists of an integer N, indicating the number of saucers. Each saucer is represented on a single line, consisting of three integers X, Y, R, indicating the coordinate and the radius. You can assume that the distance between any two saucers will never be zero.

 

Output

For each test case, please output a floating number with six fractional numbers, indicating the shortest distance.
Constraints
0 < T <= 10
2 <= N <= 50000
0 <= X, Y, R <= 100000

 

Sample Input

1 2 0 0 1 10 10 1

 

Sample Output

12.142136

 

Source

2009 Asia Wuhan Regional Contest Online