1133 non-overlapping segments (greedy algorithm, maximum interval not coincident problem)

There are n segments on the x-axis, and each segment has 1 start s and end E. The maximum number of lines that can be selected for each non-overlapping segment. (Note: The start or end point overlaps, not overlapping). For example: [1 5][2 3][3 6], you can choose [2 3][3 6], these 2 line segments do not overlap. Input

Line 1th: 1 number N, number of segments (2 <= n <= 10000) 2-n + 1 lines: 2 numbers per line, start and end of line segments ( -10^9 <= s,e <= 10^9)

Output

Outputs the maximum number of segments that can be selected.

Input example

31 52 33 6

Output example

2 
 
 
 Code as follows: 
 #include <cstdio> 
 #include <algorithm> 
 # Define MAXN 100010 
 using namespace std; 
 struct node 
 {
  int L, R; 
}; 
 BOOL CMP (Node A, Node B) 
 {
  return A.R < B.R; 
} 
 Node NUM[MAXN]; 
 Int main () 
 {
  int n; 
  while (scanf ("%d", &n)! = EOF) 
  {
   for (int i = 0; i < n; i++) 
    sca NF ("%d%d", &NUM[I].L, &NUM[I].R); 
   sort (num, num + N, CMP); 
   int ans = 0; 
   if (n>0) 
   {
    ans = 1; 
    int temp = NUM[0].R; 
    for (int i = 0; i < n;i++) 
    if (num[i].l >= temp) 
    { 
     temp = NUM[I].R; 
     ans++; 
    } 
   } 
   printf ("%d\n", ans); 
  } 
  return 0; 
} 

1133 non-overlapping segments (greedy algorithm, maximum interval not coincident problem)