BZOJ 3170: [Tjoi 2013]松鼠聚会( sort )

题目的距离为max(|x1-x2|, |y1-y2|) (切比雪夫距离). 切比雪夫距离(x, y)->曼哈顿距离((x+y)/2, (x-y)/2) (曼哈顿(x, y)->切比雪夫(x+y, x-y)). 转成Manhattan distance后排序前缀和维护即可.

--------------------------------------------------------------------------

#include<cstdio>#include<cstring>#include<algorithm> using namespace std; #define X(o) x[rx[o]]#define Y(o) y[ry[o]] typedef long long ll; const int maxn = 100009; ll smx[maxn], smy[maxn], ans;int N, x[maxn], y[maxn], rx[maxn], ry[maxn]; bool cmpX(const int &l, const int &r) { return x[l] < x[r];} bool cmpY(const int &l, const int &r) { return y[l] < y[r];} int BS(int c[], int r[], int v) { int L = 0, R = N - 1; while(L <= R) { int m = (L + R) >> 1; if(c[r[m]] == v) return m; c[r[m]] < v ? L = m + 1 : R = m - 1; } return -1;} int main() { scanf("%d", &N); for(int i = 0; i < N; i++) { rx[i] = ry[i] = i; int _x, _y; scanf("%d%d", &_x, &_y); x[i] = _x + _y; y[i] = _x - _y; } sort(rx, rx + N, cmpX); sort(ry, ry + N, cmpY); smx[0] = X(0); smy[0] = Y(0); for(int i = 1; i < N; i++) { smx[i] = smx[i - 1] + X(i); smy[i] = smy[i - 1] + Y(i); } ans = 1LL << 60; for(int i = 0; i < N; i++) { int px = BS(x, rx, x[i]), py = BS(y, ry, y[i]); ll t = 0; if(!px) { t += smx[N - 1] - ll(N) * x[i]; } else { t += ll(x[i]) * (px * 2 + 2 - N) - 2LL * smx[px] + smx[N - 1]; } if(!py) { t += smy[N - 1] - ll(N) * y[i]; } else { t += ll(y[i]) * (py * 2 + 2 - N) - 2LL * smy[py] + smy[N - 1]; } ans = min(t, ans); } printf("%lld\n", ans >> 1); return 0;}-------------------------------------------------------------------------

3170: [Tjoi 2013]松鼠聚会

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 874 Solved: 421
[Submit][Status][Discuss]

Description

有N个小松鼠，它们的家用一个点x,y表示，两个点的距离定义为：点(x,y)和它周围的8个点即上下左右四个点和对角的四个点，距离为1。现在N个松鼠要走到一个松鼠家去，求走过的最短距离。

Input

第一行给出数字N，表示有多少只小松鼠。0<=N<=10^5
下面N行，每行给出x,y表示其家的坐标。
-10^9<=x,y<=10^9

Output

表示为了聚会走的路程和最小为多少。

Sample Input

6
-4 -1
-1 -2
2 -4
0 2
0 3
5 -2

Sample Output

20

HINT

Source