NY-78-circle pool [Gift Wrapping convex hull]

Pool Time Limit: 3000 MS | memory limit: 65535 KB difficulty: 4

Description

There is a farm with many water supply devices on it. Now the farm owner wants to enclose these water supply devices with a fence to prevent their own livestock from drinking water, each pool is marked with its own coordinates. Now you need to write a program to use the shortest fence to enclose these water supply devices! (There are enough fences and the length is variable)

Input

The first line inputs N, which indicates that N groups of test data are used (1 <= N <= 10)
M is input in the second row, which indicates that the test data of this group contains m water supply devices (3 <= m <= 100)
Next, the m line represents the vertical and horizontal coordinates of each water supply device.

Output

The coordinate points of each fence passing through each water supply device are output from small to large according to the X axis coordinate value. If the X axis coordinate value is the same, then the output is from small to large according to the Y axis coordinate value.

Sample Input

140 01 12 33 0

Sample output

0 02 33 0

Code:

I just wrote a question in the same method, which is almost the same as HDU 1348 and POJ 1113.

I thought about it.

I thought the point obtained after the first sorting is ordered. What is the result? WA?

So I changed the queue storage to array storage, and sorted it once ~

# Include <stdio. h> # include <algorithm> # include <math. h> # include <string. h> using namespace std; typedef struct point {int x, y;} point; point v [105], res [105]; int vis [105], ver; int cmp (point a, point B) {if (. x <B. x) return 1; else if (. x = B. x) {if (. y <B. y) return 1; return 0;} return 0;} int calc (point a, point B, point c) // Cross Product {return (B. x-a.x) * (c. y-a.y)-(c. x-a.x) * (B. y-a.y);} double dis (point a, point B) {return s Qrt (double) (. x-b.x) * (. x-b.x) +. y-b.y) * (. y-b.y);} double solve () {vis [0] = 1; int in, I, bu, k = 0; in = 0; res [k ++] = v [in]; while (1) {bu =-1; for (I = 0; I <ver; I ++) {if (! Vis [I]) {bu = I; break;} if (bu =-1) break; for (I = 0; I <ver; I ++) {if (calc (v [in], v [bu], v [I])> 0 | (calc (v [in], v [bu], v [I]) = 0 & dis (v [in], v [I])> dis (v [in], v [bu]) bu = I;} if (vis [bu]) break; res [k ++] = v [bu]; vis [bu] = true; in = bu ;} sort (res, res + k, cmp); // here !!! For (I = 0; I <k; I ++) printf ("% d \ n", res [I]. x, res [I]. y);} int main () {int I, t; scanf ("% d", & t); while (t --) {scanf ("% d ", & ver); for (I = 0; I <ver; I ++) scanf ("% d", & v [I]. x, & v [I]. y); sort (v, v + ver, cmp); memset (vis, 0, sizeof (vis); solve ();} return 0 ;}