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題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3706
  /**//*
 題意：
給定N(N <= 100000)個點，要求找到任意一個點的最近點，點的距離定義如下：
Distance (A, B) = max (|Ax – Bx|, |Ay – By|)，求出所有點的Distance和。
解法：
二維線段樹
思路：
我的做法是利用二維線段樹的二維區間求和，首先數字太大，需要離散化，然后
將所有數字插入到離散后的二維線段樹區間，二維線段樹維護一個Sum域，表示該二維
區間插入過的數字總和，然后對于每個點，二分枚舉和它最近的點的距離，根據這個
距離找到最大的被這個距離包含的并且在二維線段樹管轄范圍的二維區間，查詢那段
區間的Sum值，如果Sum值大于1的話說明這個范圍不止一個點，于是可以縮小枚舉的范
圍，直到找到最小的枚舉值，就是這個點的最近點，總的復雜度O(nlog(n)log(n))。
這里需要注意的是，二維區間的Sum和統計和插入點的時候處理不好可能會在同一
區間插入多次或者統計多次，這和區間最值有所不同，需要特殊處理二維區間的某一維
的長度為1的情況。
 */
 #include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 200010
#define eps 1e-6
#define maxaxis 1000000000
#define ll __int64
typedef int Type;
Type bin[maxn];
int size, tot;
int B(Type val)  {
int l = 0;
int r = tot-1;
  while(l <= r) {
int m = (l + r) >> 1;
if(bin[m] == val)
return m;
if(val > bin[m]) {
l = m + 1;
 }else
r = m - 1;
}
}
struct point {
int x, y;
int _ix, _iy;
}P[maxn];
struct Tree {
int son[4];
Type Sum;
void clear() {
Sum = 0;
for(int i = 0; i < 4; i++)
son[i] = -1;
}
}T[maxn*6];
int root, all;
void Insert(int& root, int x, int y, int lx, int rx, int ly, int ry) {
if(x > rx || x < lx || y > ry || y < ly)
return ;
if(root == -1) {
root = all++;
T[root].clear();
}
if(lx == rx && x == lx && y == ly && ly == ry) {
T[root].Sum ++;
return ;
}
int midx0 = (lx + rx) >> 1;
int midy0 = (ly + ry) >> 1;
int midx1 = (lx==rx) ? midx0 : midx0 + 1;
int midy1 = (ly==ry) ? midy0 : midy0 + 1;
if(lx == rx) {
Insert(T[root].son[0], x, y, lx, midx0, ly, midy0);
Insert(T[root].son[1], x, y, lx, midx0, midy1, ry);
}else if(ly == ry) {
Insert(T[root].son[1], x, y, lx, midx0, midy1, ry);
Insert(T[root].son[2], x, y, midx1, rx, ly, midy0);
}else {
Insert(T[root].son[0], x, y, lx, midx0, ly, midy0);
Insert(T[root].son[1], x, y, lx, midx0, midy1, ry);
Insert(T[root].son[2], x, y, midx1, rx, ly, midy0);
Insert(T[root].son[3], x, y, midx1, rx, midy1, ry);
}
int i;
T[root].Sum = 0;
for(i = 0; i < 4; i++) {
if(T[root].son[i] != -1) {
T[root].Sum += T[ T[root].son[i] ].Sum;
}
}
}
void Query(int root, int x0, int x1, int y0, int y1,
int lx, int rx, int ly, int ry, Type& Num) {
if(x0 > rx || x1 < lx || y0 > ry || y1 < ly || root == -1)
return ;
if(Num > 1)
return ;
if(x0 <= lx && rx <= x1 && y0 <= ly && ry <= y1) {
Num += T[root].Sum;
return ;
}
int midx0 = (lx + rx) >> 1;
int midy0 = (ly + ry) >> 1;
int midx1 = (lx==rx) ? midx0 : midx0 + 1;
int midy1 = (ly==ry) ? midy0 : midy0 + 1;
if(lx == rx) {
Query(T[root].son[0], x0, x1, y0, y1, lx, midx0, ly, midy0, Num);
Query(T[root].son[1], x0, x1, y0, y1, lx, midx0, midy1, ry, Num);
}else if(ly == ry) {
Query(T[root].son[1], x0, x1, y0, y1, lx, midx0, midy1, ry, Num);
Query(T[root].son[2], x0, x1, y0, y1, midx1, rx, ly, midy0, Num);
}else {
Query(T[root].son[0], x0, x1, y0, y1, lx, midx0, ly, midy0, Num);
Query(T[root].son[1], x0, x1, y0, y1, lx, midx0, midy1, ry, Num);
Query(T[root].son[2], x0, x1, y0, y1, midx1, rx, ly, midy0, Num);
Query(T[root].son[3], x0, x1, y0, y1, midx1, rx, midy1, ry, Num);
}
}
int LargeThan(int val, int pre) {
int l = 0;
int r = pre;
int ans = l;
while(l <= r) {
int m = (l + r) >> 1;
if(val <= bin[m]) {
r = m - 1;
ans = m;
}else
l = m + 1;
}
return ans;
}
int LessThan(int val, int pre) {
int l = pre;
int r = tot - 1;
int ans = r;
while(l <= r) {
int m = (l + r) >> 1;
if(val >= bin[m]) {
l = m + 1;
ans = m;
}else
r = m - 1;
}
return ans;
}
int Process(int idx, int dis) {
int xleft = LargeThan( P[idx].x - dis, P[idx]._ix );
int yleft = LargeThan( P[idx].y - dis, P[idx]._iy );
int xright = LessThan( P[idx].x + dis, P[idx]._ix );
int yright = LessThan( P[idx].y + dis, P[idx]._iy );
int Num = 0;
Query(root, xleft, xright, yleft, yright, 0, tot-1, 0, tot-1, Num);
return Num;
}
int n;
int main() {
int i;
int Max;
while(scanf("%d", &n) != EOF) {
size = 0;
tot = 0;
all = 0;
Max = 0;
for(i = 0; i < n; i++) {
scanf("%d %d", &P[i].x, &P[i].y);
bin[ size++ ] = P[i].x;
bin[ size++ ] = P[i].y;
if(P[i].x > Max) Max = P[i].x;
if(P[i].y > Max) Max = P[i].y;
}
sort(bin, bin + size);
for(i = 0; i < size; i++) {
if(i == 0 || bin[i] != bin[i-1])
bin[tot++] = bin[i];
}
root = -1;
for(i = 0; i < n; i++) {
P[i]._ix = B(P[i].x);
P[i]._iy = B(P[i].y);
Insert(root, P[i]._ix, P[i]._iy, 0, tot-1, 0, tot-1);
}
ll ans = 0;
for(i = 0; i < n; i++) {
int l = 0;
int r = Max;
int dis = 0;
while(l <= r) {
int m = (l + r) >> 1;
if(Process(i, m) > 1) {
r = m - 1;
dis = m;
}else
l = m + 1;
}
ans += dis;
}
printf("%I64d\n", ans);
}
return 0;
}
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