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  1. #include<cstdio>
  2. #include<cstring>
  3. #include<iostream>
  4. using namespace std;
  5. #define MAX 1010
  6. #define MAXLVL 12
  7. #define rl(x) scanf("%d",&x)
  8. int parent[MAX+12],dp[MAXLVL+2][MAX],depth[MAXLVL+12];
  9. int LCA(int u, int v) 
  10. {
  11. 	if(depth[u] < depth[v]) 
  12. 	swap(u,v);
  13. 	int diff = depth[u] - depth[v];
  14. 	for(int i=0; i<MAXLVL; i++) 
  15. 	if( (diff>>i)&1 ) 
  16. 	{
  17. 		u = dp[i][u];
  18. 	}
  19. 	if(u == v) 
  20. 	return u;
  21. 	for(int i=MAXLVL-1; i>=0; i--) 
  22. 	if(dp[i][u] != dp[i][v]) 
  23. 	{
  24. 		u = dp[i][u];
  25. 		v = dp[i][v];
  26. 	}
  27. 	return dp[0][u];
  28. }
  29. int main()
  30. {
  31. 	int t,ca=1;
  32. 	rl(t);
  33. 	while(ca<=t)
  34. 	{
  35. 		int n,m;
  36. 		rl(n);
  37. 		memset(depth,0,sizeof depth);
  38. 		parent[0]=-1;
  39. 		for(int i=0;i<n;i++)
  40. 		{
  41. 			rl(m);
  42. 			while(m--)
  43. 			{
  44. 				int x,y;
  45. 				rl(y);
  46. 				y--;
  47. 				depth[y]=depth[i]+1;
  48. 				parent[y]=i;
  49. 			}
  50. 		}
  51. 		for(int i=0;i<MAXLVL;i++)
  52. 		    for(int j=0; j<n; j++)
  53. 			dp[i][j]=-1;
  54. 		for(int i=0;i<n;i++)
  55. 			dp[0][i]=parent[i];
  56. 		for(int i=1; i<MAXLVL; i++)
  57. 			for(int j=0; j<n; j++)
  58. 				if(dp[i-1][j] != -1)
  59. 					dp[i][j] = dp[i-1][dp[i-1][j]];
  60. 		printf("Case %d:\n",ca);
  61. 		int q,x,y;
  62. 		rl(q);
  63. 		while(q--)
  64. 		{
  65. 			cin>>x>>y;
  66. 			x--,y--;
  67. 			cout<<LCA(x,y)+1<<endl;
  68. 		}
  69. 		ca++;
  70.  
  71. 	}
  72. }
Success #stdin #stdout 0s 2792KB
comments (?)
stdin
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1
7
3 2 3 4
0
3 5 6 7
0
0
0
0
2
5 7
2 7
stdout
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Case 1:
3
1
https://ideone.com/b2igzl
language:
C++ 4.3.2 (gcc-4.3.2)
created:
9 years ago
visibility:
public

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