UVA 11012 - Cosmic Cabbages（枚举技巧）

Problem A

Cosmic Cabbages 

Input: Standard Input
Output: Standard Output
 

CABBAGE, n. A familiar kitchen-garden vegetable about as large and wise as a man's head.

Ambrose Bierce
Scientists from the planet Zeelich have figured out a way to grow cabbages in space. They have constructed a huge 3-dimensional steel grid upon which they plant said cabbages. Each cabbage is attached to a corner in
the grid, where 6 steel cables meet and is assigned Cartesian coordinates. A cosmic ant wants to crawl from cabbage X to cabbage Y along the cables that make the grid. The cosmic ant always chooses the shortest possible path along the grid lines while going
from cabbage X to cabbage Y. This distance is called the cosmic distance between two cabbages. Given a collection of cabbages what is the maximum distance between any two of the cabbages?

Input
The first line of input gives the number of cases, N (0<N<21). N test cases follow. Each one starts with a line containing n (2<=n<=105). The next n lines will
each give the 3-dimensional coordinates of a cosmic cabbage (integers in the range [-108, 108]).

 

Output

For each test case, output one line containing "Case #x:" followed by the largest cosmic distance between cabbages X and Y, out of all possible choices of X and Y.

 

Sample Input                               Output for Sample Input

4 2 1 1 1 2 2 2 3 0 0 0 0 0 1 1 1 0 4 0 1 2 3 4 5 6 7 8 9 10 11 6 0 0 0 1 1 1 2 2 2 0 0 1 1 0 0 0 1 0

Case #1: 3 Case #2: 3 Case #3: 27 Case #4: 6

[align=center][/align]

题意：给定n个三维坐标点，求两两最大曼哈顿距离。

思路：直接枚举O（n^2）不可行，那么换个方法：列出公式 d = |x1 - x2| + |y1 - y2| + |z1 - z2|。对应8种情况这里不一一列举，就举其中x1 - x2 + y1 - y2 + z1 - z2.。转换为(x1 + y1 + z1) - (x2 + y2 + z2)其他同理。发现只要前面尽可能大，后面尽可能小，出来的曼哈顿距离必然最大，那么对应8种情况，每种对于每个点都枚举过去，保存下所有点中x +|- y +|- z 最大和最小的值，最大的作为减数，最小最为被减数，然后对应8种情况中求出最大的即可。

代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define INF 0x3f3f3f3f
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)

const int N = 100005;

int t, n;
struct Point {
int v[3];
}p
;

void init() {
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d%d%d", &p[i].v[0], &p[i].v[1], &p[i].v[2]);
}

int solve() {
int ans = 0;
for (int i = 0; i < (1<<3); i++) {
int Min = INF, Max = -INF;
for (int j = 0; j < n; j ++) {
int sum = 0;
for (int k = 2; k >= 0; k--) {
if (i&(1<<k)) sum += p[j].v[2 - k];
else sum -= p[j].v[2 - k];
}
Max = max(Max, sum); Min = min(Min, sum);
}
ans = max(ans, Max - Min);
}
return ans;
}

int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
init();
printf("Case #%d: %d\n", ++cas, solve());
}
return 0;
}