AcWing 119. 袭击 分治

标签：typedef   return   include   algo   const   坐标   参考   turn   struct   

参考博客：https://www.acwing.com/solution/AcWing/content/826/

//#include<bits/stdc++.h>
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100100;
const int INF=1e7;
struct node
{
	double x,y;
	int f;
}a[N<<1],t[N<<1];
int n,m;
int cmp(node a,node b)
{
	if(a.x==b.x)
		return a.y<b.y;
	return a.x<b.x;
}
int cmp2(node a,node b)
{
	if(a.y==b.y)
		return a.x<b.x;
	return a.y<b.y;
} 
double dist(node a,node b)
{
	if(a.f==b.f)
		return INF;
	double dx=(a.x-b.x),dy=(a.y-b.y);
	return sqrt(dx*dx+dy*dy);
}
double solve(int l,int r)
{
	//先把坐标按x排序
	//然后取终点
	//左边答案是ans1，右边答案是ans2
	//那么答案就是，
	//min(ans1,ans2,两边交叉)
	//对于两边，直接递归
	int mid=(l+r)>>1;
    if (r-l==0)
        return INF;
    if (r-l==1)
        return dist(a[l],a[r]);
    double ans=min(solve(l,mid),solve(mid+1,r));
    //交叉部分 
    int cnt=0;
    for (int i=l;i<=r;i++)
        if (a[i].x>=a[mid].x-ans && a[i].x<=a[mid].x+ans)
            t[++cnt]=a[i];
    sort(t+1,t+1+cnt,cmp2);
    for (int i=1;i<=cnt;i++)
        for (int j=i+1;j<=cnt;j++)
        {
            if(t[j].y>=t[i].y+ans)
                break;
            ans=min(ans,dist(t[i],t[j]));
        }
    return ans;
}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		cin>>n;
		for(int i=1;i<=n;i++)
		{
			cin>>a[i].x>>a[i].y;
			a[i].f=0;
		} 
		for(int i=n+1;i<= n<<1;i++)
		{
			cin>>a[i].x>>a[i].y;
			a[i].f=1;
		}
		n<<=1;
		sort(a+1,a+1+n,cmp);
		printf("%0.3lf\n",solve(1,n));
	}
}

AcWing 119. 袭击 分治

标签：typedef   return   include   algo   const   坐标   参考   turn   struct   

原文地址：https://www.cnblogs.com/QingyuYYYYY/p/12638298.html