La 3695 distant galaxy

The coordinates of N points are given (the coordinates are all positive numbers), and the maximum number of points can be on the boundary of a rectangle.

In the solution, several arrays are constructed, which are displayed in the figure.

First, two horizontal lines are enumerated. Then left [I] indicates the point on the left of the vertical line I, and on [I] indicates the two horizontal lines on the vertical line I (not on the horizontal line) the number of points. on2 [I] indicates the number of points that are located between the two horizontal lines on the vertical line I plus the horizontal line boundary.

So the points on the rectangle frame are:

Left [J]-left [I] + on [I] + on2 [J]

Enumerate the right boundary vertical line J. After J is determined, maintain the maximum value of on [I]-left [I.

 

1 // # define local 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 using namespace STD; 6 7 const int maxn = 100 + 10; 8 int n, m, Y [maxn], on [maxn], on2 [maxn], left [maxn]; 9 10 struct point11 {12 int x, Y; 13 bool operator <(const point & RHs) const14 {15 return x <RHS. x; 16} 17} p [maxn]; 18 19 int solve () 20 {21 sort (p, p + n); 22 sort (Y, Y + n ); 23 m = unique (Y, Y + n)-y; // m is different from Y coordinate 24 if (M <= 2) 25 return N; 26 27 int ans = 0; 28 for (int A = 0; A <m; ++) 29 for (int B = a + 1; B <m; ++ B) 30 {31 int ymin = Y [a], Ymax = Y [B]; 32 33 // calculate left, on, on234 int K = 0; // K record the number of bars 35 for (INT I = 0; I <n; ++ I) 36 {37 if (I = 0 | P [I]. x! = P [I-1]. x) 38 {// This is a new vertical line 39 + + k; 40 on [k] = on2 [k] = 0; 41 left [k] = k = 0? 0: left [k-1] + on2 [k-1]-on [k-1]; 42} 43 If (P [I]. y> ymin & P [I]. Y <Ymax) 44 + + on [k]; 45 if (P [I]. y> = ymin & P [I]. Y <= Ymax) 46 + + on2 [k]; 47} 48 if (k <= 2) 49 return N; 50 51 int m = 0; 52 for (Int J = 1; j <= K; ++ J) 53 {54 ans = max (ANS, left [J] + on2 [J] + M ); 55 m = max (M, on [J]-left [J]); 56} 57} 58 59 return ans; 60} 61 62 int main (void) 63 {64 # ifdef local65 freopen ("rj5in.txt", "r", stdin); 66 # endif67 68 int Kase = 0; 69 while (scanf ("% d ", & n) = 1 & N) 70 {71 for (INT I = 0; I <n; ++ I) 72 {73 scanf ("% d ", & P [I]. x, & P [I]. y); 74 y [I] = P [I]. y; 75} 76 printf ("case % d: % d \ n", ++ Kase, solve (); 77} 78 return 0; 79}

Code Jun

 

Conclusion: The question in the big white book is wonderful, and you cannot fully understand the essence of the question after careful consideration. It must be because I am too weak to answer questions.