该题具有一定的技巧性,需要在Nlog(N)的时间复杂度下计算出任意一个点,N-1个点到其的距离综和,这里需要运用这样一个技巧,将x,y分开计算,首先计算x轴的距离,那么就先排一次序,然后有到P号点的距离和为 (P-1) * Xp - sum(X1...Xp-1) + sum(Xp+1...Xn) - (N-P) * xp; 同理可计算出y轴的距离,这两个距离是累加到一个结构体上的。所以最后直接找最小值即可。
代码如下:
#include <cstdlib> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; typedef long long int Int64; int N; struct Point { Int64 x, y, sum; }e[100005]; bool cmpx(Point a, Point b) { return a.x < b.x; } bool cmpy(Point a, Point b) { return a.y < b.y; } int main() { int T; Int64 sum, Min; scanf("%d", &T); while (T--) { sum = 0; Min = 1LL << 62; scanf("%d", &N); for (int i = 1; i <= N; ++i) { e[i].sum = 0; scanf("%I64d %I64d", &e[i].x, &e[i].y); } sort(e+1, e+1+N, cmpx); for (int i = 1; i <= N; ++i) { e[i].sum += (i-1) * e[i].x - sum; sum += e[i].x; } sum = 0; for (int i = N; i >= 1; --i) { e[i].sum += sum - (N-i) * e[i].x; sum += e[i].x; } sum = 0; sort(e+1, e+1+N, cmpy); for (int i = 1; i <= N; ++i) { e[i].sum += (i-1) * e[i].y -sum; sum += e[i].y; } sum = 0; for (int i = N; i >= 1; --i) { e[i].sum += sum - (N-i) * e[i].y; Min = min(Min, e[i].sum); sum += e[i].y; } printf("%I64d\n", Min); } return 0; }