hdu 4311

题意

平面上\(n(n\leq 1e5)\)个点，找一个点到其它所有点的曼哈顿距离之和最小。

思路

如果是找一个坐标使得所有点到其曼哈顿距离之和最小，那么将\(n\)个横坐标排个序，取中间的一个为答案的横坐标，将\(n\)个纵坐标排个序，取中间的一个为答案的纵坐标。原因就是绝对值$$y=|x-a_1|+|x-a_2|+...+|x-a_n|$$的图像为平底锅型或者是尖底。因为可以在平面上任意取点，所以可以取最优的\(x\)和\(y\).

但是这道题并不能够任意取点，而是限定在了\(n\)个点中。怎么办呢？

最常规的想法，就是将距离都算出来，取个最小值。然而直接算的话是\(O(n^2)\)的，数据量显然不允许。那么就换种算的方法。

还是先排序，考虑序列\(a_1,a_2,...,a_n\)（已升序排好），则$$|a_i-a_1|+|a_i-a_2|+...+|a_i-a_{i-1}|+|a_i-a_{i+1}|+...+|a_i-a_n|$$$$=((a_i-a_1)+(a_i-a_2)+...+(a_i-a_{i-1}))+((a_{i+1}-a_i)+...+(a_n-a_i))$$$$=(i-1)a_i-\sum_{k=1}^{{i-1}a_k-(n-i)*a_i+\sum_{k=i+1}}{n}a_k$$$$=(2i-1-n)*a_i-\sum_{k=1}^{{i-1}a_k+\sum_{k=i+1}}{n}a_k$$

于是可以预处理前缀和后缀和，就可以在\(O(n)\)的时间处理出来各个点对应的值了。

最后每个点的横坐标距离和纵坐标距离加起来取个最小值即可。

算法复杂度\(O(nlogn)\).

Code

#include <bits/stdc++.h>

#define inf 0x3f3f3f3f3f3f3f3f

#define maxn 100010

using namespace std;

typedef long long LL;

struct node {

    LL x, y;

}a[maxn];

bool cmp1(int i, int j) { return a[i].x < a[j].x; }

bool cmp2(int i, int j) { return a[i].y < a[j].y; }

LL x[maxn], y[maxn], pr[maxn], su[maxn];

int id[maxn];

void work() {

    int n;

    scanf("%d", &n);

    for (int i = 1; i <= n; ++i) scanf("%lld%lld", &a[i].x, &a[i].y);

    for (int i = 1; i <= n; ++i) id[i] = i;

    sort(id+1, id+1+n, cmp1);

    memset(pr, 0, sizeof pr);

    memset(su, 0, sizeof su);

    for (int i = 1; i <= n; ++i) pr[i] = pr[i-1] + a[id[i]].x;

    su[n] = a[id[n]].x; for (int i = n-1; i > 0; --i) su[i] = su[i+1] + a[id[i]].x;

    for (int i = 1; i <= n; ++i) x[id[i]] = (2*i-1-n)*a[id[i]].x - pr[i-1] + su[i+1];

    sort(id+1, id+1+n, cmp2);

    memset(pr, 0, sizeof pr);

    memset(su, 0, sizeof su);

    for (int i = 1; i <= n; ++i) pr[i] = pr[i-1] + a[id[i]].y;

    su[n] = a[id[n]].y; for (int i = n-1; i > 0; --i) su[i] = su[i+1] + a[id[i]].y;

    for (int i = 1; i <= n; ++i) y[id[i]] = (2*i-1-n)*a[id[i]].y - pr[i-1] + su[i+1];

    LL ans = inf;

    for (int i = 1; i <= n; ++i) ans = min(ans, x[i]+y[i]);

    printf("%lld\n", ans);

}

int main() {

    int T;

    scanf("%d", &T);

    while (T--) work();

    return 0;

}

hdu 4312

题意

平面上\(n(n\leq 1e5)\)个点，找一个点到其它所有点的切比雪夫距离之和最小。

思路

将切比雪夫距离转化为曼哈顿距离，方法为将坐标转45度。即将\((x,y)\)的坐标映射为\((x+y,y-x)\).

然后就可以直接套上一题了，最终答案除以2.

Code

#include <bits/stdc++.h>

#define inf 0x3f3f3f3f3f3f3f3f

#define maxn 100010

using namespace std;

typedef long long LL;

struct node {

    LL x, y;

}a[maxn];

bool cmp1(int i, int j) { return a[i].x < a[j].x; }

bool cmp2(int i, int j) { return a[i].y < a[j].y; }

LL x[maxn], y[maxn], pr[maxn], su[maxn];

int id[maxn];

void work() {

    int n;

    scanf("%d", &n);

    for (int i = 1; i <= n; ++i) {

        LL xx, yy;

        scanf("%lld%lld", &xx, &yy);

        a[i].x = xx+yy, a[i].y = yy-xx;

    }

    for (int i = 1; i <= n; ++i) id[i] = i;

    sort(id+1, id+1+n, cmp1);

    memset(pr, 0, sizeof pr);

    memset(su, 0, sizeof su);

    for (int i = 1; i <= n; ++i) pr[i] = pr[i-1] + a[id[i]].x;

    su[n] = a[id[n]].x; for (int i = n-1; i > 0; --i) su[i] = su[i+1] + a[id[i]].x;

    for (int i = 1; i <= n; ++i) x[id[i]] = (2*i-1-n)*a[id[i]].x - pr[i-1] + su[i+1];

    sort(id+1, id+1+n, cmp2);

    memset(pr, 0, sizeof pr);

    memset(su, 0, sizeof su);

    for (int i = 1; i <= n; ++i) pr[i] = pr[i-1] + a[id[i]].y;

    su[n] = a[id[n]].y; for (int i = n-1; i > 0; --i) su[i] = su[i+1] + a[id[i]].y;

    for (int i = 1; i <= n; ++i) y[id[i]] = (2*i-1-n)*a[id[i]].y - pr[i-1] + su[i+1];

    LL ans = inf;

    for (int i = 1; i <= n; ++i) ans = min(ans, x[i]+y[i]);

    printf("%lld\n", ans/2);

}

int main() {

    int T;

    scanf("%d", &T);

    while (T--) work();

    return 0;

}