枚举三角形，然后判断是否其余的点都不在形内，也不在边上。时间复杂度是O(n
4)。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-9-16 20:29:12
File Name: pku1569.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{ for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) { for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
typedef struct point_t
{
char name;
int x, y;
};
point_t p[20];
point_t operator -(const point_t &a, const point_t &b)
{
point_t ret;
ret.x = a.x - b.x;
ret.y = a.y - b.y;
return ret;
}
int cross(const point_t &a, const point_t &b)
{
return a.x * b.y - a.y * b.x;
}
int area(int a, int b, int c)
{
return cross(p[b] - p[a], p[c] - p[a]);
}
int inside(int t, int a, int b, int c)
{
int ab = area(t, a, b), bc = area(t, b, c), ca = area(t, c, a);
if (ab >= 0 && bc >= 0 && ca >= 0 || ab <= 0 && bc <= 0 && ca <= 0)
return 1;
else return 0;
}
int main()
{
int n, ans;
int ans_a, ans_b, ans_c;
while (scanf("%d\n", &n), n != 0)
{
for (int i = 0; i < n; i++)
scanf("%c%d%d\n", &p[i].name, &p[i].x, &p[i].y);
ans = 0;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
for (int k = j + 1; k < n; k++)
{
int flag = 1;
for (int t = 0; t < n && flag; t++)
if (t != i && t != j && t != k)
if (inside(t, i, j, k))
flag = 0;
if (flag)
{
int t = abs(area(i, j, k));
if (t > ans)
{
ans = t;
ans_a = i;
ans_b = j;
ans_c = k;
}
}
}
printf("%c%c%c\n", p[ans_a].name, p[ans_b].name, p[ans_c].name);
}
return 0;
}