HDU-4311 Meeting point-1 曼哈顿距离快速计算

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;

}