CF 600 D explanations

CF600D problem solution

The title literally means

       Gives an undirected graph, this requirement is harmonious FIG harmony defines that if A to C (A <B <C) of a road, then the B to C is also a way. A data range is 2e5

Then we realized that if all the elements of a subset is continuous, then this subset is harmony, because the connected graph, any two points can be reached. If a subset of the elements is not continuous, then you certainly need to have a multiple paths or even go to another subset. The two examples given illustrate:

 

 

 

 

 

Then it becomes a question of how to deal with.

I use the disjoint-set, for the point of attachment, are incorporated into the small to the point, and then update the array maxx array size (size [i] is used to record the number i is set at this time, to record set maxx the maximum)

Then we can judge very pleasant, if maxx [i] - i + 1 == size [i], then that is the harmony of this collection, that is, do not need to connect to the outside edge.

Then for a condition is not satisfied how to deal with it?

Very simple traversal i + 1 to maxx [i], of the need to merge the incoming set of combined can, in this process, maxx [i] are always updated, the complexity of this idea is O (n), very elegant.

 Finally put rookie Code

 1 #include <stdio.h>
 2 #include <algorithm>
 3 using namespace std;
 4 const int N = 2e5+10;
 5 int fa[N], maxx[N], size[N];
 6 int find(int x) //并查集基本操作 
 7 {
 8     if(x==fa[x]) return x;
 9     return fa[x] = find(fa[x]);
10 }
11 int main()
12 {
13     int x, y, fx, fy;
14     int n, m, res = 0;
15     scanf("%d %d", &n, &m);
16     for(int i = 1; i <= n; i++)  //并查集的初始化 
17     {
18         fa[i] = i;
19         maxx[i] = i;
20         size[i] = 1;
21     }
22     for(int i = 1; i <= m; i++)
23     {
24         scanf("%d %d", &x, &y);
25         //判断 然后建里并查集 
26         fx = find(x); fy = find(y);
27         if(fx < fy)
28         {
29             fa[fy] = fx;
30             size[fx] += size[fy];
31             maxx[fx] = max(maxx[fx], maxx[fy]);
32         }else if(fy < fx){
33             fa[fx] = fy;
34             size[fy] += size[fx];
35             maxx[fy] =max (Maxx [FX], Maxx [FY]);
 36          }
 37 [      }
 38 is      for ( int I = . 1 ; I <= n-;)
 39      {
 40          IF (FA [I] == I) // this is determined not to be is a   
41          {
 42              // satisfies the condition 
43 is              IF (size [I] == Maxx [I] -i + . 1 ) + I = size [I];
 44 is              the else {
 45                  // not satisfy the condition 
46 is                  for ( int J = + I . 1 ; J <= Maxx [I]; J ++ )
 47                 {
 48                      // dynamically updating answer ++ 
49                      IF (Find (J) =! I)
 50                      {
 51 is                          FA [J] = I;
 52 is                          size [I] + = size [J];
 53 is                          Maxx [I] = max ( Maxx [I], Maxx [J]);
 54 is                          RES ++ ;
 55                      }
 56 is                  }
 57 is              }
 58          }
 59      }
 60      the printf ( " % D \ n- " , RES);
61     
62 }