题目传送门:
点这传送
题目大意:
给你n个整数点,现在让你找一个点,使得该点到这些点的距离之和最小,问这样子的点有多少个?
题目思路:
如果改题目是一维线段,将数组从小到大排序,如果n是奇数,则最短点唯一,若n是偶数,则最短点有两个
现在题目扩展为二维平面,只需要将x坐标和y坐标分别从小到大排序,若n为奇数,结果唯一,若为偶数,则求交叉面积就可以(因为题目说了所建点可以和整数点重合)
代码:
#include<bits/stdc++.h>
// #include <iostream>
// #include <stdio.h>
// #include <math.h>
// #include <string.h>
// #include <iomanip>
// #include <vector>
// #include <queue>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define endl '\n'
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long int ll;
typedef __int64 bi;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){
return x&(-x);}
ll gcd(ll a,ll b){
return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){
if(!b){
d=a,x=1,y=0;}else{
ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){
ll res=1;a%=MOD;while(b>0){
if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){
return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){
ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){
ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {
if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();return x*f; }
int dir[4][2] = {
{
1,0}, {
-1,0},{
0,1},{
0,-1} };
int x[1002],y[1002];
signed main(void)
{
int t,n;
t = read();
while(t--){
n = read();
for(int i=1;i<=n;++i){
x[i] = read(); y[i] = read();
}
sort(x+1,x+n+1);
sort(y+1,y+n+1);
ll midx = x[n/2+1] - x[n/2] + 1;
ll midy = y[n/2+1] - y[n/2] + 1;
if(n&1) cout<<1<<endl;
else printf("%lld\n",midx * midy);
}
}