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最新推荐文章于 2019-09-25 19:34:18 发布

最新推荐文章于 2019-09-25 19:34:18 发布 · 121 阅读

本文深入探讨了排序算法如何在O(n)时间复杂度内进行枚举，详细介绍了各种排序方法和其实现技巧，适用于计算机科学领域的专业人士。

排序后用O（n）时间枚举

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;

#ifdef _WIN32
typedef __int64 i64;
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
typedef long long i64;
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FFF(i,a)        for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const int maxn = 100011;

struct zz
{
	i64 x;
	i64 y;
	int id;
}zx;
i64 n;
i64 fx[maxn];
i64 fy[maxn];
i64 sum[maxn];

inline bool cmpx(const zz & a,const zz & b) 
{
	return a.x<b.x;
}
	 
inline bool cmpy(const zz & a,const zz & b) 
{
	return a.y<b.y;
}

vector<zz>v;

void findx()
{
	sort(v.begin(),v.end(),cmpx);
	i64 temp;
	for(int i=1;i<v.size();i++)
	{
		fx[i] += fx[i-1];
		temp = v[i].x-v[i-1].x;	
		temp *= i;
		fx[i] += temp;
		sum[v[i].id]+=fx[i];
	}	
	for(int i=1;i<=n;i++)
	{
		fx[i]=0;
	}
	int cnt = v.size()-1;
	for(int i=cnt-1;i>=0;i--)
	{
		fx[i] += fx[i+1];
		temp = v[i+1].x-v[i].x;
		temp *= (cnt - i);	
		fx[i] += temp;		
		sum[v[i].id]+=fx[i];
	}

	return ;
}

void findy()
{
	sort(v.begin(),v.end(),cmpy);
	i64 temp;
	for(int i=1;i<v.size();i++)
	{
		fy[i] += fy[i-1];
		temp = v[i].y-v[i-1].y;	
		temp *= i;
		fy[i] += temp;
		sum[v[i].id]+=fy[i];
	}	
	for(int i=1;i<=n;i++)
	{
		fy[i]=0;
	}
	int cnt = v.size()-1;
	for(int i=cnt-1;i>=0;i--)
	{
		fy[i] += fy[i+1];
		temp = v[i+1].y-v[i].y;
		temp *= (cnt - i);	
		fy[i] += temp;		
		sum[v[i].id]+=fy[i];
	}
	return ;
}

i64 start()
{
	findx();
	findy();
	i64 ans = inf64;
	for(int i=1;i<=n;i++)
	{
		if(sum[i]<ans)
		{
			ans=sum[i];
		}
	}
	return ans;
}

int main()
{
	int T;
	cin>>T;		
	while(T--)
	{		
		cin>>n;
		v.clear();
		for(int i=0;i<=n;i++)
		{
			fx[i]=0;
			fy[i]=0;
			sum[i]=0;
		}
		for(int i=1;i<=n;i++)
		{	
			cin>>zx.x>>zx.y;
			zx.id = i;
			v.pb(zx);
		}	
		cout<<start()<<endl;
	}
	return 0;
}