UVa 1312 - Cricket Field (高效枚举)

题意

找出最大的正方形，不包含点。

思路

先仰慕帆神。

枚举左右边界，然后从下到上枚举上边界，维护最大值和左下角的坐标。

代码

  
  
  
  

   
   
   
   
int n, vis[MAXN], width;
   
   
   
   
vector<int> x;
   
   
   
   
pii P[MAXN];
   
   
   
   

   
   
   
   
bool cmp(const pii &a, const pii &b)
   
   
   
   
{
   
   
   
   
 return a.Y < b.Y;
   
   
   
   
}
   
   
   
   

   
   
   
   
void Handle()
   
   
   
   
{
   
   
   
   
 for (int i = 0; i < n; i++) scanf("%d%d", &P[i].X, &P[i].Y);
   
   
   
   
 x.clear();
   
   
   
   
 MS(vis, 0);
   
   
   
   
 vis[0] = 1;
   
   
   
   
 x.PB(0);
   
   
   
   
 for (int i = 0; i < n; i++)
   
   
   
   
 {
   
   
   
   
 if (!vis[P[i].X])
   
   
   
   
 {
   
   
   
   
 vis[P[i].X] = 1;
   
   
   
   
 x.PB(P[i].X);
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 if (!vis[width]) x.PB(width);
   
   
   
   
 sort(x.begin(), x.end());
   
   
   
   
}
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 //ROP;
   
   
   
   
 int height, T;
   
   
   
   
 scanf("%d", &T);
   
   
   
   
 while (T--)
   
   
   
   
 {
   
   
   
   
 scanf("%d%d%d", &n, &width, &height);
   
   
   
   
 Handle();
   
   
   
   
 sort(P, P + n, cmp);
   
   
   
   
 int maxValue = 0, curBottom;
   
   
   
   
 pii ans;
   
   
   
   
 for (int i = 0; i < SZ(x); i++)
   
   
   
   
 {
   
   
   
   
 for (int j = i+1; j < SZ(x); j++)
   
   
   
   
 {
   
   
   
   
 int tmpWide = x[j] - x[i], curBottom = 0;
   
   
   
   
 for (int k = 0; k < n; k++)
   
   
   
   
 {
   
   
   
   
 if (P[k].X <= x[i] || P[k].X >= x[j]) continue;
   
   
   
   
 int tmpHeight = P[k].Y - curBottom;
   
   
   
   
 if (maxValue < min(tmpWide, tmpHeight))
   
   
   
   
 {
   
   
   
   
 maxValue = min(tmpWide, tmpHeight);
   
   
   
   
 ans.X = x[i]; ans.Y = curBottom;
   
   
   
   
 }
   
   
   
   
 curBottom = P[k].Y;
   
   
   
   
 }
   
   
   
   
 int tmpHeight = height - curBottom;
   
   
   
   
 if (maxValue < min(tmpWide, tmpHeight))
   
   
   
   
 {
   
   
   
   
 maxValue = min(tmpWide, tmpHeight);
   
   
   
   
 ans.X = x[i]; ans.Y = curBottom;
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 printf("%d %d %d\n", ans.X, ans.Y, maxValue);
   
   
   
   
 if (T) puts("");
   
   
   
   
 }
   
   
   
   
 return 0;
   
   
   
   
}