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\[|a_i-a_1|+|a_i-a_2|+...+|a_i-a_{i-1}|+|a_i-a_{i+1}|+...+|a_i-a_n|\]\[=((a_i-a_1)+(a_i-a_2)+...+(a_i-a_{i-1}))+((a_{i+1}-a_i)+...+(a_n-a_i))\]\[=(i-1)*a_i-\sum_{k=1}^{i-1}a_k-(n-i)*a_i+\sum_{k=i+1}^{n}a_k\]\[=(2*i-1-n)*a_i-\sum_{k=1}^{i-1}a_k+\sum_{k=i+1}^{n}a_k\]
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
#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
#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;
}