传送门
B. Eastern Exhibition
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You and your friends live in n houses. Each house is located on a 2D plane, in a point with integer coordinates. There might be different houses located in the same point. The mayor of the city is asking you for places for the building of the Eastern exhibition. You have to find the number of places (points with integer coordinates), so that the summary distance from all the houses to the exhibition is minimal. The exhibition can be built in the same point as some house. The distance between two points (x1,y1) and (x2,y2) is |x1−x2|+|y1−y2|, where |x| is the absolute value of x.
Input
First line contains a single integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains a single integer n (1≤n≤1000). Next n lines describe the positions of the houses (xi,yi) (0≤xi,yi≤109).
It’s guaranteed that the sum of all n does not exceed 1000.
Output
For each test case output a single integer - the number of different positions for the exhibition. The exhibition can be built in the same point as some house.
思路
分别寻找x和y坐标中的中位数，若n为奇数，则展览馆建在中位数坐标处，数目恒为1；若n为偶数，则可建在左中位数坐标到右中位数坐标之间（包括这两个点）的所有点。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll x[1010],y[1010];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(x,0,sizeof(x));
memset(y,0,sizeof(y));
int n;
scanf("%d",&n);
for(int i = 1; i <= n; i++)
{
scanf("%lld",&x[i]);
scanf("%lld",&y[i]);
}
sort(x+1, x+1+n);
sort(y+1, y+1+n);
if(n%2)
{
printf("1\n");
}
else
{
ll lx = x[(n+1)/2],ly = y[(n+1)/2];
ll rx = x[(n+1)/2+1],ry = y[(n+1)/2+1];
ll ans = (rx - lx + 1)*(ry - ly + 1);
if(ans == 0)
printf("1\n");
else
printf("%lld\n",ans);
}
}
}