Problem G. Garden Gathering Input file: standard input Output file: standard output Time limit: 3 seconds Memory limit: 512 megabytes Many of you may have been to St. Petersburg, but have you visited Peterhof Palace? It is a collection of splendid palaces and gardens with spectacular fountains! Besides the beauty, it is huge, and you can easily get lost in one of the park labyrinths. Imagine that you are not a regular visitor, but one of the guides, and your group of tourists is scattered across one of the gardens — a complete disaster! To continue the tour, you need to collect them all in one place, and technologies of the XXI century could be very useful in this task. Each tourist has a smartphone with a GPS tracker which transmits data directly to your phone. Unfortunately, the application for Peterhof’s guides lacks in functionality. Actually, it has the only button which, when pressed, automatically selects one person at random and tells his or her coordinates to everyone in the group. After that, all tourists immediately start to move to this position using the shortest path, while the selected person stands still and waits for others. The only thing to worry about is that you can be late for the last train home, so you want to know the maximum possible time this gathering process could take. You have a map of this garden with you: -5 -5 -4 -4 -3 -3 -2 -2 -1 -1 0 1 1 2 2 3 3 4 4 5 5 Ox Oy Picture 1: Plan of garden trails Page 9 of 17 ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 All tourists from your group travel through the park with constant speed using only the trails shown on the picture 1. If in the end you will be late, then you can ask your boss to reimburse the money spent on Yandex.Taxi. To do so, you need to present a proof in the form of two numbers: ID of the person selected by the app and ID of the person who will be the last to arrive. As you have a lot of time while the tourists are gathering, calculate any possible pair for the worst case. Input The first line of the input contains a single integer n (2 ≤ n ≤ 200 000) — the size of your group. The i-th of the next n lines contains two integers xi and yi (|xi |, |yi | ≤ 107 ) — coordinates of the tourist with ID i (numbered from 1 to n). Initial positions of all tourists are guaranteed to be distinct. Output Output ID of the selected person and ID of the last person. If there are several possible answers, output any of them. Examples standard input standard output 4 0 0 2 0 0 1 2 1 1 4 Note Picture 2: Answer for the first sample In the sample the distance between the first and the fourth tourists is √ 2 + 1. Answers (4, 1), (2, 3), and (3, 2) are also considered correct.
题意：给出方格图上N个点，点都在格点上，每个点都可以沿着上下左右，斜着四个方向，总共八个方向移动，时间花费为线的长度，问N个点到其中一个点集中，每个点都沿着最短路移动，问最后到的点最晚的时候，起点和最后到的点是哪两个点。（任意一组答案）。
分析：显然答案就是最远点对。
所以我们要求曼哈顿距离下的最远点对。
按照曼哈顿生成树的思想来做。
八个方向，根据双向性，其实每个点只用处理四个方向。而这四个方向不妨选择上面的四个方向，这样我们可以通过坐标变换用相同的方法处理。
所以我们就只用考虑一个方向，不妨考虑y=0 到 y=x 这两条直线围成的那个方向。
首先，设现在这个点为(x0, y0)，要找一个这个方向上的最远点。
这个点既然在这个方向上，首先就要有x1>x0，并且y1-y0<x1-x0
然后观察时间的表达式 time = sqrt(2)*y + (x-y)
所以我们现在要求的是，对于每个点(x0, y0)，找出所有的x1>x0,y1-y0<x1-x0，询问它们之中的max{time}
要保证x1>x0，我们可以将x坐标排序，然后倒序计算点即可。
要保证y1-y0<x1-x0,变形
y1-x1<y0-x0，
即找到比y0-x0小的所有yi-xi，然后找它们的max{time}，这个可以用树状数组做。
注意的是：我觉得，以吾之拙见，上述方法可能将不是这个方向上的点算进来，
因为如果x0,y0 -> (0, 0)，x1,y1 -> (2, -3)这样的话，虽然不在这个方向上，
但是x1>x0，y1-y0<x1-x0，
虽然可能算多，但是这种方法是不可能将这个方向上的点算少的。
也就是说，在这个方向上的最大值必定有被考虑到。
所以不影响答案和复杂度。
每个点四个方向，即上述过程要做四次，所以要O(4*nlogn)
/**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define ft first
#define sd second
#define mk make_pair inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = ;
const DB SQRT2 = sqrt(2.0);
class Point
{
// private : static const double AAAAA = sqrt(234);
public :
int x, y, index;
inline void Read()
{
x = Getint();
y = Getint();
} inline bool operator <(const Point &a) const
{
if(x != a.x) return x < a.x;
return y < a.y;
} static DB Dist(const Point &a, const Point &b)
{
int dx = abs(a.x - b.x), dy = abs(a.y - b.y);
if(dx < dy) swap(dx, dy);
return SQRT2 * dy + (dx - dy);
} inline const DB Feature()
{
return SQRT2 * y + (x - y);
}
} data[N];
int n;
int st, far;
DB ans;
class Hash
{
private :
int arr[N], length; public :
inline void Clear()
{
length = ;
} inline void Insert(int x)
{
arr[++length] = x;
} inline void GoHash()
{
sort(arr + , arr + + length);
length = unique(arr + , arr + + length) - (arr + );
//for(int i = 1; i <= length; i++) printf("%d\n", arr[i]);
} inline int GetIndex(int x)
{
int left = , right = length, mid;
while(left <= right)
{
mid = (left + right) >> ;
if(arr[mid] < x) left = mid + ;
else if(arr[mid] > x) right = mid - ;
else return mid;
}
return ;
} inline int GetLength()
{
return length;
}
} ranks; class TreeArray
{
private :
DB value[N];
int index[N], n; public :
inline void Clear()
{
n = ;
} inline void SetLimit(int x)
{
n = x;
for(int i = ; i <= n; i++) value[i] = -1.0 * INF;
} inline int Lowbit(int x)
{
return x & (-x);
} inline void Add(Point a)
{
int x = ranks.GetIndex(a.y - a.x);
//printf("%d %d %d\n", a.y - a.x, x, n);
DB val = a.Feature();
for( ; x <= n; x += Lowbit(x))
if(val > value[x])
value[x] = val, index[x] = a.index;
} inline void Query(const Point &a, DB &cnt, int &idx)
{
cnt = -1.0 * INF, idx = ;
int x = ranks.GetIndex(a.y - a.x);
for( ; x; x -= Lowbit(x))
if(cnt < value[x])
cnt = value[x], idx = index[x];
}
} Store; inline void Input()
{
n = Getint();
for(int i = ; i <= n; i++)
{
data[i].Read();
data[i].index = i;
}
} inline void Solve()
{
//puts("adf");
ans = -1.0 * INF, st = far = ;
for(int dir = ; dir < ; dir++)
{
//puts("asdfff");
if(dir == || dir == )
{
for(int i = ; i <= n; i++)
swap(data[i].x, data[i].y);
}
else if(dir == )
{
for(int i = ; i <= n; i++)
data[i].x = -data[i].x;
} ranks.Clear();
for(int i = ; i <= n; i++)
ranks.Insert(data[i].y - data[i].x);
ranks.GoHash(); sort(data + , data + + n);
for(int i = ; i <= n; i++)
Store.Clear();
Store.SetLimit(ranks.GetLength());
//puts("asdfffffx");
for(int i = n; i >= ; i--)
{
//printf("%d", i);
DB cnt;
int idx;
Store.Query(data[i], cnt, idx);
//puts("xxx");
if(idx)
{
cnt -= data[i].Feature();
if(cnt > ans) ans = cnt, st = data[i].index, far = idx;
}
Store.Add(data[i]);
//puts("yyy");
}
} printf("%d %d\n", st, far);
} int main()
{
Input();
Solve();
return ;
}
有人能够解答为何在类里面
private : static const double AAAAA = sqrt(234.0);
不可以吗？