hdu 4311 排序+二分

枚举每个点是必须的，问题就在于如何快速求出曼哈顿距离之和。可以将x坐标和y坐标分别排序，枚举到点(xx,yy)的时候在排好序的x和y数组中二分找到其位置，然后之前的数字都小于，之后的数字都大于，就可以去掉绝对值了。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 using namespace std;
 6
 7 typedef long long ll;
 8 const ll INF = 9999999999999999;
 9 const int N = 100000;
10 int x[N];
11 int y[N];
12 ll sumx[N];
13 ll sumy[N];
14 ll ans;
15
16 struct Point
17 {
18     int x, y;
19 } point[N];
20
21 int main ()
22 {
23     int t;
24     scanf("%d", &t);
25     while ( t-- )
26     {
27         int n;
28         scanf("%d", &n);
29         for ( int i = 0; i < n; i++ )
30         {
31             scanf("%d", x + i);
32             scanf("%d", y + i);
33             point[i].x = x[i];
34             point[i].y = y[i];
35         }
36         sort( x, x + n );
37         sort( y, y + n );
38         sumx[0] = x[0];
39         sumy[0] = y[0];
40         for ( int i = 1; i < n; i++ )
41         {
42             sumx[i] = sumx[i - 1] + x[i];
43             sumy[i] = sumy[i - 1] + y[i];
44         }
45         ans = INF;
46         for ( int i = 0; i < n; i++ )
47         {
48             int xx = point[i].x;
49             int yy = point[i].y;
50             int posx = upper_bound( x, x + n, xx ) - x;
51             int posy = upper_bound( y, y + n, yy ) - y;
52             ll tmpx = ( ll ) posx * xx - sumx[posx - 1] + sumx[n - 1]
53             - sumx[posx - 1] - ( ll ) ( n - posx ) * xx;
54             ll tmpy = ( ll ) posy * yy - sumy[posy - 1] + sumy[n - 1]
55             - sumy[posy - 1] - ( ll ) ( n - posy ) * yy;
56             if ( tmpx + tmpy < ans )
57             {
58                 ans = tmpx + tmpy;
59             }
60         }
61         printf("%I64d\n", ans);
62     }
63     return 0;
64 }