[ACM-ICPC] Disjoint Set

Transcript

1. Disjoint Set ֲࢸݢʢKuoE0ʣ KuoE0.tw@gmail.com KuoE0.ch
2. Latest update: Mar 13, 2013 Attribution-ShareAlike 3.0 Unported (CC BY-SA 3.0) http://creativecommons.org/licenses/by-sa/3.0/
3. Disjoint Set தᅳᩄɿฒ查ू 1 4 3 7 8 5 2 6 9
4. Disjoint Set தᅳᩄɿฒ查ू 1 4 3 7 8 5 2 6 9
5. Disjoint Set தᅳᩄɿฒ查ू 1 4 3 7 8 5 2 6 9
6. Disjoint Set தᅳᩄɿฒ查ू 1 4 3 7 8 5 2 6 9
7. Tree Structure 1 5 2 6 4 3 7 8 9 1
8. Tree Structure 1 5 2 6 4 3 7 8 9
9. Data Structure index 0 1 2 3 4 content parent node parent node parent node parent node parent node The parent of root is itself!
10. Find Operation 1 5 2 6 node parent 1 1 2 1 5 2 6 2
11. 1 5 2 6 How to determine that node 2 and node 6 in the same set?
12. 1 5 2 6 How to determine that node 2 and node 6 in the same set? 6 2
13. 1 5 2 6 How to determine that node 2 and node 6 in the same set? 6 2 2 1
14. 1 5 2 6 How to determine that node 2 and node 6 in the same set? 6 2 2 1 1
15. Source Code // parent is an array stored the parent of nodes. int findSet( int x ) { if ( parent[ x ] == x ) return x; return findSet( parent[ x ] ); }
16. ... ... ... ... If the height of tree is very large... ... ... } n
17. Reduce Height 1 5 2 6
18. Reduce Height 1 5 2 6 1 5 2 6
19. Reduce Height node parent 1 1 2 1 5 1 6 1 1 5 2 6
20. Source Code // parent is an array stored the parent of nodes. int findSet( int x ) { if ( parent[ x ] == x ) return x; return parent[ x ] = findSet( parent[ x ] ); }
21. Union Operation 1 5 2 6 7 8 9 Height = 3 Height = 2
22. Union Operation 1 5 2 6 7 8 9 Height = 3 Height = 2
23. Union Operation 1 5 2 6 7 8 9 Height = 3
24. Source Code // parent is an array stored the parent of nodes. // height is an array stored the height of tree void unionSet( int a, int b ) { if ( a == b ) return; if ( height[ a ] > height[ b ] ) parent[ b ] = a; else { parent[ a ] = b; if ( height[ a ] == height[ b ] ) ++height[ b ]; } }
25. Time Complexity Find Operation O(1) Union Operation O(1)
26. Practice Now [UVa] 10583 - Ubiquitous Religions
27. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
28. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
29. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
30. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
31. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
32. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
33. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
34. 10583 - Ubiquitous Religions ဠڭ ဠڭ ဠڭ ဠڭ ဠڭ جಜڭ جಜڭ ճڭ جಜڭ
35. Source Code #include <iostream> #include <cstdio> using namespace std; #define MAXN 50010 int parent[ MAXN ], height[ MAXN ]; int main() { int n, t = 0, e, a, b; while ( scanf( “%d %d”, &n, &e ) && n && e ) { for ( int i = 1; i <= n; ++i ) parent[ i ] = i; height[ i ] = 1; for ( int i = 0; i < e; ++i ) { scanf( “%d %d”, &a, &b ); unionSet( findSet( a ), findSet( b ) ); } int ret = 0; for ( int i = 1; i <= n; ++i ) if ( parent[ i ] == i ) ++ret; printf( “Case %d: %d\n”, ++t, ret ); } return 0; }
36. Practice Now [UVa] 10608 - Friends
37. Thank You for Your Listening.